Answer:
D) A group 1 alkali metal bonded to chloride, such as RbCl
Explanation:
Which of the following has the greatest electronegativity difference between the bonded atoms?
A) A strong acid made of hydrogen and a halogen, such as HCl
The electronegativity difference between H and Cl is:
ΔEN = |EN(H)-EN(Cl)| = |2.1-3.0| = 0.9
B) A diatomic gas, such as nitrogen (N₂)
The electronegativity difference between N and N is:
ΔEN = |EN(N)-EN(N)| = |3.0-3.0| = 0.0
C) Carbon bonded to a Group 6A (16) nonmetal chalcogen, such as in CO
The electronegativity difference between C and O is:
ΔEN = |EN(C)-EN(O)| = |2.5-3.5| = 1.0
D) A group 1 alkali metal bonded to chloride, such as RbCl
The electronegativity difference between Rb and Cl is:
ΔEN = |EN(Rb)-EN(Cl)| = |0.8-3.0| = 2.2
Why are atoms of carbon-14(C-14) unstable?
. What is the difference between the geometries of the bonding of the two oxygen atoms in acetic acid
Answer:
One oxygen atom is tetrahedral while the other is trigonal planar
Explanation:
Acetic acid contains two oxygen atoms. The two oxygen atoms are different and have different geometries.
The first oxygen atom which is part of the carbonyl moiety in the molecule is in a trigonal planar geometry since it is sp2 hybridized.
The second oxygen atom is in sp3 hybridized state hence it is tetrahedral.
Thus one oxygen atom is tetrahedral and the other is trigonal planar in acetic acid.
A standard galvanic cell is constructed in which a Cu^2+ | Cu^+ half cell acts as the cathode Which of the following statements are correct?
(Choose all that apply)
A. Cr^3+|Cr could be the other standard half cell.
B. As the cell runs, anions will migrate from the other compartment to the Cu^2+|Cu^+ compartment.
C. I_2|I^- could be the other standard half cell.
D. Cu^+ is oxidized at the cathode.
E. In the external circuit, elections flow from the other compartment to the Cu^2+|Cu^+ compartment
Answer:
E. In the external circuit, elections flow from the other compartment to the Cu^2+|Cu^+ compartment
Explanation:
In a galvanic cell, electrons flow from anode to cathode. We must remember that oxidation occurs at the anode and reduction occurs at the cathode. Hence the process; M(s) -------> M^+(aq) + e occurs at the anode.
The electrons lost at the anode are conveyed to the cathode where they are accepted by other chemical species and are reduced to the corresponding reduced species according to the reaction; M^+(aq) + e -----> M(s)
Hence the Cu^2+ | Cu^+ half cell, being the cathode accepts electrons from the other half cell for this reduction reaction to take place, hence the answer
Acetic acid and water react to form hydronium cation and acetate anion, like this: (aq)(l)(aq)(aq) Imagine of are added to a flask containing a mixture of , , and at equilibrium, and then answer the following questions. What is the rate of the forward reaction before any HCH3CO2 has been added to the flask
The question is incomplete. Here is the complete question.
Acetic acid and water react to form hydronium cation and acetate anion, like this:
[tex]HCH_{3}CO_{2}_{(aq)}+H_{2}O_{(l)} -> H_{3}O^{+}_{(aq)}+CH_{3}CO_{2}^{-}_{(aq)}[/tex]
Imagine 226 mmol of [tex]CH_{3}CO_{2}^{-}[/tex] are added to a flask containing a mixture of [tex]HCH_{3}CO_{2}[/tex], [tex]H_{2}O[/tex],[tex]H_{3}O^{+}[/tex] and [tex]CH_{3}CO_{2}^{-}[/tex] at equilibrium and then answer the following questions:
1) What's the rate of the reverse reaction before any [tex]CH_{3}CO_{2}^{-}[/tex] has been added to the flask?
a) 0
b) Greater than 0, but less than the rate of the forward reaction
c) Greater than 0, but equal to the rate of the forward reaction
d) Greater than 0, but greater than the rate of the forward reaction
2) What is the rate of the reverse reaction just after the [tex]CH_{3}CO_{2}^{-}[/tex] has been added to the flask?
a) 0
b) Greater than 0, but less than the rate of the forward reaction
c) Greater than 0, but equal to the rate of the forward reaction
d) Greater than 0, but greater than the rate of the forward reaction
3) What is the rate of the reverse reaction when the system has again reached equilibrium?
a) 0
b) Greater than 0, but less than the rate of the forward reaction
c) Greater than 0, but equal to the rate of the forward reaction
d) Greater than 0, but greater than the rate of the forward reaction
4) How much more [tex]CH_{3}CO_{2}^{-}[/tex] is in the flask when the system has again reached equilibrium?
a) None
b) Some, but less than 226 mmol
c) 226 mmol
d) More than 226 mmol.
Answer: 1) c) Greater than 0, but equal to the rate of the forward reaction
2) d) Greater than 0, but greater than the rate of the forward reaction
3) c) Greater than 0, but equal to the rate of the forward reaction
4) b) Some, but less than 226 mmol
Explanation: A reversible chemical reaction reaches its equilibrium when forward and reverse reaction are at the same rate. At that point, equilibrium has a constant called K.
Equilibrium constant depends on the concentration of its products and reagents.
For the reaction [tex]HCH_{3}CO_{2}_{(aq)}+H_{2}O_{(l)} -> H_{3}O^{+}_{(aq)}+CH_{3}CO_{2}^{-}_{(aq)}[/tex],
the forward is towards production of [tex]CH_{3}CO_{2}^{-}[/tex] and reverse is towards the production of [tex]HCH_{3}CO_{2}[/tex].
1) When equilibrium is reached, forward and reverse are at the same rate and are different from zero.
2) When adding a compound, the equilibrium is broken. So, to go back to the equilibrium, reaction tend to counteract the change. In the case of acetic acid and water above, when adding acetate anion, the reverse reaction will produce more acetic acid to restore equilibrium, so reverse reaction will be at a greater rate and different from 0.
3) After a while when the system is back to the equilibrium, the rate will be equal again.
4) After second equilibrium, acetate anion will have less mmol than when this new equilibrium state started.
Which diagram shows a pattern similar to the emission spectrum of hydrogen?
Arrange the following substances in the order of increasing entropy at 25°C.
a. HF(g)
b. NaF(s)
c. SiF 4(g)
d. SiH 4(g)
e. Al(s)
Answer:
In solving this question it is important to note that entropy is referred to as the degree of randomness in a compound or system.
Entropy is highest in gas and lowest in solids
The order is given as:
Gas > Liquid > Solid
Another criteria is the more the elements in a compound, the more the entropy.
For example: KO3 has more entropy than KO2 due to the presence of more elements.
The last criteria is calculating their masses , the one with a greater mass has the higher entropy.
The option with the least entropy will be a solid. Al(s) and NaF(s) are both solids. NaF(s) has more elements in a compound which makes it have a higher entropy than Al(s). This means NaF(s) > Al(s).
The next to consider are the gaseous compounds which are HF(g) , SiF4(g) and SiH4(g). The least entropy will be HF(g) . This is because it has the least number of elements of the 3 compounds.
SiF4(g) and SiH4(g) have equal number of elements in a compound but the one with the highest mass is SiF4(g) because fluorine has a higher mass when compared to hydrogen. This means the compound with the highest entropy is SiF4(g).
The order of increasing entropy is given below:
Al(s) < NaF(s) < HF(g) < SiH4(g) < SiF4(g)
Calculate the freezing point of a solution of 500.0 g of ethylene glycol dissovled in 500g water. Kf = 1.86 degrees C/m and Kb (which my instructor said was just extraneous info that is not used here) is 0.512 degrees C/m.
Answer: [tex]T_{f}[/tex] = -29.96 °C
Explanation: A solution has a lower freezing point compared to the pure solvent, due to the solute lowering the vapor pressure of the solvent.
The "new" freezing point is calculated as:
[tex]\Delta T_{f} = K_{f}.m[/tex]
[tex]K_{f}[/tex] is the molal freezing-point depression constant
m is molality concentration, i.e., moles of solute per kilogram of solvent
Before calculating freezing point, let's find moles of ethylene glycol ([tex]C_{2}H_{6}O_{2}[/tex]) in the solution:
Molar mass [tex]C_{2}H_{6}O_{2}[/tex] = 62.08 g/mol
For 500 g:
[tex]n = \frac{500}{62.08}[/tex]
n = 8.05 moles
The molality concentration for 0.5kg of water:
m = [tex]\frac{8.05}{0.5}[/tex]
m = 16.11
The freezing point will be:
[tex]\Delta T_{f} = K_{f}.m[/tex]
[tex]\Delta T_{f} = -1.86*16.11[/tex]
[tex]\Delta T_{f} = -29.96[/tex]
Freezing point of a solution of Ethylene Glycol and Water is -29.96°C
Deduce the identity of the following compound from the spectral data given. C3H4BrN: H NMR, δ 2.98 (2H, triplet), 3.53 (2H, triplet) 13C NMR, δ 21.05 (triplet), 23.87 (triplet), 118.08 (singlet) (ppm):JR, 2963, 2254 cm" Express your answer as a chemical formula.
Answer: Br-CH₂-CH₂-C≡N (3-bromopropanenitrile)
Explanation:
The question tells us to " Deduce the identity of the following compound from the spectral data given. C3H4BrN: H NMR, δ 2.98 (2H, triplet), 3.53 (2H, triplet) 13C NMR, δ 21.05 (triplet), 23.87 (triplet), 118.08 (singlet) (ppm):JR, 2963, 2254 cm" Express your answer as a chemical formula".
Answer: Another Alternative to this is seen below
The Double Bond Equivalent of C₃H₄BrN = 3+1-4/2-1/2+1/2 = 2
2254 cm-1 Vc ≡N [IR analysis]
2963.4 = sp³Vc-H streek [IR analysis]
N≡C-CH₂-CH₂-Br (C₃H₄NBr)where the first CH₂ = b = 2.98 (2H,triplet) and
the second CH₂ = a = 3.53 (2H,triplet)
Note: where both are 1HNMR.
N≡C-CH₂-CH₂-Br
where C = c = 118.08 (singlet)
CH₂ = b = 21.05 (triplet)
CH₂ = a = 23.87 ( triplet)
Note: where all are 13CNMR
Important:
I have attached a copy of the solution to enhance better understanding in case the typed solution isn't clear enough.
A 0.0447−mol sample of a nutrient substance, with a formula weight of 114 g/mol, is burned in a bomb calorimeter containing 6.19 × 102 g H2O. Given that the fuel value is 6.13 × 10−1 in nutritional Cal when the temperature of the water is increased by 5.05°C, what is the fuel value in kJ in scientific notation?
Answer:
The value is [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]
Explanation:
From the question we are told that
The no of moles of the sample is n = 0.0447 mole
The formula weight is [tex]M = 114 \ g/mol[/tex]
The mass of water is [tex]m = 6.19 *10^{2}\ g[/tex]
The amount of the fuel is [tex]f= 6.13*10^{-1} \ nutritional \ Cal[/tex]
The temperature rise is [tex]\Delta T = 5.05^o[/tex]
Generally
[tex]1 \ nutritional \ Cal => 4.184*10^{3} \ kJ/kg[/tex]
=> [tex]f= 6.13*10^{-1} \ nutritional \ Cal \to x[/tex]
=> [tex]x = \frac{6.13 *10^{-1} * 4.184 *10^{3}}{1}[/tex]
=> [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]
9.
The main difference between the salt in the Dead
Sea and salt in the school lab is that
A salt in the Dead Sea is saltier.
B salt in the Dead Sea is iodized while that
of the lab is uniodized.
C salt in the Dead Sea is sodium chloride while
that of the lab has its hydrogen replaced.
D salt in the Dead Sea is sodium chloride
while salt in the lab is either soluble or
insoluble in water.
The correct option is A.
What is the main difference between salt in the Dead Sea and salt in the school lab?
Ordinary sea salt is 97 percent sodium chloride whereas Dead Sea salt is a mixture of lots of different chloride and bromide salts.
What's the difference between sea salt and Dead Sea salt?Dead Sea Salt differs greatly from other sea salts in mineral content, is made up of sodium chloride with a high percentage of magnesium, sulfates, and potassium. You can bring that ancient, therapeutic experience home with our guaranteed authentic Dead Sea bath salts.
Learn more about salt here https://brainly.com/question/2626194
#SPJ2
he mass of a single molybdenum atom is 1.59×10-22 grams. How many molybdenum atoms would there be in 38.3 milligrams of molybdenum?
Answer:
2.41 × 10²⁰ atoms
Explanation:
Step 1: Given data
Mass of a single molybdenum atom: 1.59 × 10⁻²² gMass of the molybdenum sample: 38.3 mgStep 2: Convert the mass of the sample to grams
We will use the relationship 1 g = 1,000 mg.
38.3 mg × (1 g/1,000 mg) = 0.0383 g
Step 3: Calculate the number of atoms in 0.0383 g of molybdenum
0.0383 g × (1 atom/1.59 × 10⁻²² g) = 2.41 × 10²⁰ atom
What is the chemical potential of an element in its standard state? Select the correct answer below: −1.00kJ/mol 0kJ/mol 1.00kJ/mol 9.384kJ/mol
Answer:
0 kJ/mol.
Explanation:
Hello,
In this case, since the chemical potential can be represented in terms of the Gibbs free energy of formation:
[tex]\mu = \frac{\Delta _fG}{n}[/tex]
Thus, since the Gibbs free energy of formation of an element is zero, the chemical potential is also zero, or just 0 kJ/mol.
Best regards.
Need help I need a short imaginary science lab story where several safety rules were ignored by students and things went wrong include at least 4 safety rules that were ignored.
A proposed mechanism for the reaction: 2 H2 + 2NO - N2 + 2H30 Step 1: H2g) + 2NO () - N2O (g) + H20) (slow) Step 2: N20cm) + H2 (g) - N2 (g) + H2O) (fast) What is the rate law and the intermediate respectively
Answer:
[tex]\text{Rate} = k\, [\mathrm{H_2\, (g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].
[tex]\rm N_2O\, (g)[/tex] is the intermediate.
Explanation:
RateBalanced overall reaction: [tex]\rm 2\, H_2\, (g) + 2\, NO\, (g) \to N_2\, (g) + 2\, H_2O\, (g)[/tex].
Proposed mechanism:
[tex]\rm H_2\, (g) + 2\, NO\, (g) \to N_2O\, (g) + H_2O \, (g)[/tex] (slow.)[tex]\rm N_2O\, (g) + H_2\, (g) \to N_2\, (g) + H_2O\, (g)[/tex] (fast.)Start with the slowest, rate-determining step of the proposed mechanism. Here, the rate-determining step is [tex]\rm H_2\, (g) + 2\, NO\, (g) \to N_2O\, (g) + H_2O \, (g)[/tex].
There are two species on the reactant side of this intermediate reaction: [tex]\rm H_2\, (g)[/tex] and [tex]\rm NO_2\, (g)[/tex]. The concentration of both of them should be in the rate expression of this step.
On the other hand, the coefficient of [tex]\rm H_2\, (g)[/tex] is one while the coefficient of [tex]\rm NO_2\, (g)[/tex] is two. Therefore, in the rate expression of this step, the concentration of [tex]\rm H_2\, (g)\![/tex] should have a power of one, while the concentration of [tex]\rm NO_2\, (g)\![/tex] should have a power of two.
include the rate constant [tex]k[/tex] to obtain the rate expression of the rate-determining slow step:
[tex]\text{Rate} = k\, [\mathrm{H_2\,(g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].
Make sure that all species in this rate expression are on the reactant side of the overall balanced reaction. Otherwise, further steps would be required to obtain the rate law of the overall reaction.
Therefore, by this proposed mechanism, the rate law of the overall reaction would be [tex]\text{Rate} = k\, [\mathrm{H_2\,(g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].
IntermediateIn a proposed reaction mechanism, a species is an intermediate if it appeared in one of the proposed steps, but not in the balanced overall equation.
The two steps of this proposed mechanism mentioned five species:
[tex]\rm H_2\, (g)[/tex].[tex]\rm NO\,(g)[/tex].[tex]\rm N_2O\, (g)[/tex].[tex]\rm H_2O\, (g)[/tex].[tex]\rm N_2\, (g)[/tex].With the exception of [tex]\rm N_2O\, (g)[/tex], all the other species appeared in the overall balanced equation. Therefore, [tex]\rm N_2O\, (g)\![/tex] is the intermediate.
Draw the Lewis structure for H2O. How many lone pairs of electrons are there in the central atom?
Answer:
2 lone pairs of electrons
Explanation:
In this case, the central atom is oxygen. If we check the periodic table we will have an atomic number of 8. So, the electronic configuration would be:
[tex]1s^2~2s^2~2p^4[/tex]
We have 6 electrons in the last level. Therefore, we have 6 valence electrons. For the hydrogens, we will have only 1 electron each. So, in total, we have 8 electrons.
With this in mind, we have to draw two lone pairs in the oxygen and 1 bond with each hydrogen.
See figure 1.
I hope it helps!
Which two philosophers challenged Democritus?
Answer:
I don't know the second one but one of them is Aristotle
Answer:
Aristrotle and John Dalton
Explanation:
The reaction system POBr3(g) =POBr(g) + Brz(g) is at equilibrium. Which of the following statements describes the behavior of the system if POBr is added to the container?
A.POBr will be consumed in order to establish a new equilibrium.
B. the partial pressures of POBrand POBr will remain steady while the partial pressure of bromine increases.
C. the partial pressure of bromine will increase while the partial pressure of POBr decreases.
D. the partial pressure ofbromine remains steady while the partial pressures of POBrand POBr increase.
E, the forward reaction will proceed to establish equilibrium.
Answer:
A. POBr will be consumed in order to establish a new equilibrium.
Explanation:
POBr3(g) =POBr(g) + Brz(g)
The question is based on Le Chatelier's principle. This principle states that whenever a system at equilibrium is stressed, the system would undergo changes to annul that stress.
If more POBr is added to the system, it means that the concentration of the products have been increased and as a result the equilibrium has been disturbed. To restore equilibrium, the system would have to shift in a direction that reduces the product concentration.
This direction is the backward direction and in doing so, POBr will be consumed.
The correct option is option A.
The statement that describes the behavior of the system is:
A. POBr will be consumed in order to establish a new equilibrium.
Reaction system:POBr₃(g) ⇄ POBr(g) + Br₂(g)
According to Le Ch-atelier principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish an equilibrium.
If more POBr is added to the system, it means that the concentration of the products have been increased and as a result the equilibrium has been disturbed. To restore equilibrium, the system would have to shift in a direction that reduces the product concentration.
This direction is the backward direction and in doing so, POBr will be consumed.
Thus, the correct option is option A.
Find more information about Equilibrium here:
brainly.com/question/12270624
Larissa needs to make 500ml of 2.00 M NaCl. The molecular weight of NaCl is 58.44. How does she make 500ml of 2.00 M NaCl? Question 3 options: A) Measure 29.22g NaCl and dissolve it into 500ml water. B) Measure 29.22g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask. C) Measure 58.44g NaCl and dissolve it into 500ml of water. D) Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.
Answer:
D) Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.
Explanation:
Step 1: Given data
Molarity (M): 2.00 M
Volume (V): 500 mL = 0.500 L
Molar mass of NaCl: 58.44 g/mol
Step 2: Calculate the required moles of NaCl
We will use the following expression.
n = M × V
n = 2.00 mol/L × 0.500 L
n = 1.00 mol
Step 3: Calculate the mass corresponding to 1.00 moles of NaCl
1.00 mol × 58.44 g/mol = 58.44 g
Step 4: Describe the procedure to prepare the solution
Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.
Answer:
D) Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.
Explanation:
I got it right in class!
Hope this helps!! :))
A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the water? Group of answer choices
Answer:
33.85°C
Explanation:
From the question,
Heat lost by the hotter water = heat gained by the colder water
cm'(t₂-t₃) = cm(t₃-t₁)................. Equation 1
Where c = specific heat capacity of water, m' = mass of hot water, m = mass of cold water, t₁ = Initial temperature of cold water, t₂ = Initial temperature of hot water, t₃ = final temperature of the mixture.
But since the density of water is constant, and mass varies directly as volume, We can replace the mass of water with the volume of water. i.e,
cv'(t₂-t₃) = cv(t₃-t₁)................. Equation 2
Where v' and v are the volume of hot water and cold water respectively
make t₃ the subject of the equation
t₃ = (v't₂+vt₁)/(v'+v)............ Equation 3
Given: v' = 5.0 L, v = 60 L, t₁ = 30°C, t₂ = 80°C
Substitute these values into equation 3
t₃ = (5×80+60×30)/(60+5)
t₃ = 2200/68
t₃ = 33.85°C
why are step by step written procedures are an essential part of any scientific experiment
Answer: The scientific method attempts to minimize the influence of bias or prejudice in the experimenter. Even the best-intentioned scientists can't escape bias. ... That's the job of the scientific method. It provides an objective, standardized approach to conducting experiments and, in doing so, improves their results.
How many TOTAL ions would be produced as aqueous species in the water solvent if four molecules of aluminum perchlorate dissolved? The answer is _____ total ions
Answer:
the total I think would be 35
Phosphoric acid, H3PO4, will undergo three successive ionization reactions to varying extents in water. What is the balanced equilibrium identified as Ka3
Answer:
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺ Ka3
Explanation:
Phosphoric acid, H3PO4 is a dyprotic acid which undergoes in these three succesive ionization reactions.
The last equillibrium is conditionated to the Ka3
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ Ka1
H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺ Ka2
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺ Ka3
It is an acid because it release a proton, to make the [H₃O⁺] rise.
pH of solution will always be < 7
H₂PO₄⁻ and HPO₄⁻² are amphoteric compounds, which means that they can work as an acid (release protons), or base (take protons)
The melting point of water is 0°C at 1 atm pressure because under these conditions:
A. ΔS for the process H2O(s) → H2O(l) is positive.
B. ΔS and ΔSsurr for the process H2O(s) → H2O(l) are both positive.
C. ΔS and ΔSsurr for the process H2O(s) → H2O(l) are equal in magnitude and opposite in sign.
D. ΔG is positive for the process H2O(s) → H2O(l).
E. None of these is correct.
Answer:
The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.
Explanation:
The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:
H2O (s) ⇒ H2O (l)
The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:
Entropy change (ΔSsys) = ΔSproduct - ΔSreactant
= (69.9 - 47.89) J mol/K
= 22.0 J mol/K
Therefore, the value of entropy change is positive.
Now the value of entropy for surrounding ΔSsurr will be,
ΔSsurr = -ΔHfusion/T
= -6012 j/mol/273
= -22.0 J/molK
Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.
Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . What mass of ammonium p
Answer:
7.5 g
Explanation:
There is some info missing. I think this is the original question.
Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.
Step 1: Write the balanced equation
H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄
Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid
The molar mass of phosphoric acid is 98.00 g/mol.
[tex]4.9 g \times \frac{1mol}{98.00g} = 0.050mol[/tex]
Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid
The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.
Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate
The molar mass of ammonium phosphate is 149.09 g/mol.
[tex]0.050mol \times \frac{149.09 g}{mol} = 7.5 g[/tex]
Barium fluoride, , is partially soluble with Ksp of 1.7e-06. What is the solubility of barium fluoride in pure water
Answer:
[tex]7.5x10^{-3}M[/tex]
Explanation:
Hello,
In this case, since the dissociation barium fluoride is represented at equilibrium by:
[tex]BaF_2(s)\rightleftharpoons Ba^{2+}(aq)+2F^-(aq)[/tex]
Hence, the equilibrium expression is:
[tex]Ksp=[Ba^{2+}][F^-]^2[/tex]
Whereas the molar solubility is represented as the reaction extent [tex]x[/tex]:
[tex]Ksp=[x][2x]^2[/tex]
In such a way, we can solve for [tex]x[/tex]:
[tex]1.7x10^{-6}=4x^3\\\\x=\sqrt[3]{\frac{1.7x10^{-6}}{4} } \\\\x=7.5x10^{-3}M[/tex]
Which as said before, is the molar solubility.
Best regards.
Answer quick please I’m begging you
Answer:
The sun warms up parts of the oceans. Warm waters rise just like warm air rises. So, as the warmer ocean waters begin to rise in a particular area, the cooler ocean waters from a different area will move in to replace the warmer ocean waters, and this creates our ocean currents.
Hope it helps!
Nuclear fusion always results in
A. a net loss of mass
B. a net gain of mass
C. no change in mass (mass is conserved)
D. depends on the nucleons that fuse
Answer:
D
Explanation:
In this time there are made a lot of nuclear bombs so itatters on how what how much damage it makes according to its capacity in my opinion.
Nuclear reaction of which nuclear fusion is a type which depends on the nucleons that fuse as the total mass of single nucleus is less than that of two original nuclei.
What are nuclear reactions?There are two types of nuclear reactions which are nuclear fusion and nuclear fission .They involve the combination and disintegration of the element's nucleus respectively.
In nuclear fission, the nucleus of the atom is bombarded with electrons of low energy which splits the nucleus in to two parts .Large amount of energy is released in the process.It is used in nuclear power reactors as it produces large amount of energy.
In nuclear fusion,on the other hand, is a reaction which occurs when two or more atoms combine to form a heavy nucleus.Large amount of energy is released in the process which is greater than that of the energy which is released in nuclear fission process.
Learn more about nuclear reactions,here:
https://brainly.com/question/12786977
#SPJ5
Which of the following could be considered a scientific statement? (2 points)
Answer:
The answer of the scientific STATEMENT is" Appears That Ants Live In Colonies."
What cell would you use to electroplate silver over on iron
Answer:
You would use an electrolytic cell.
A student determines the mass of a metal rod to be 39.35 g. it is places in 10.00 mL of water, and the water level rises to 15.00 mL. What is the density of the rod?
Answer:[tex]\rho=7.87g/mL[/tex]
Explanation:
Hello,
In this case, we are analyzing a problem in which the Archimedes principle is considered, thus, as the volume of the metal rod is determined by the difference between the mass of the water and the mass of the water and the rod:
[tex]V_{rod}=15.00mL-10.00mL=5.00mL[/tex]
In such a way, the density turns out:
[tex]\rho =\frac{m_{rod}}{V_{rod}}=\frac{39.35g}{5.00mL} \\ \\\rho=7.87g/mL[/tex]
Regards.