Answer:
Explanation:
1 Mole of Aluminum with mass 26.98g contains 6.02*10^23 atoms.
In 2.88g of Aluminum, there are 2.88/26.98*6.02*10^23 = 6.426*10^22 atoms.
Answer:
Explanation:
no of Al atoms = 2.88/26.98*6.02*10^23
= 6.426*10^22
= 6.43*10^22
Write a complete list of steps you will utilize to predict the electron-domain geometry for a given species using only the Lewis structure. The only information you are provided with is the molecular formula and the net charge.
Answer:
See explanation
Explanation:
According to the valence shell electron pair repulsion theory, the shape of a molecule depends on the number of electron pairs on the valence shell of the central atom in the molecule.
So, in order to identify the electron domain geometry of a molecule (number of regions of electron density)
1) draw the full Lewis structure of the molecule/ion
2) indicate the electron pairs surrounding the central atom
3) count the number of electron pairs surrounding the central atom
4) decide on the electron pair geometry using valence shell electron pair repulsion theory.
In this exercise we have to use our knowledge of electrons to describe our molecular formula, so we have to:
The valence shell is the outermost shell that an atom can present, that is, it is the shell furthest from the nucleus of an atom.
What is the constitution of electrons?Electrons are negatively charged particles that revolve around the atomic nucleus and have a mass 1836 times less than that of protons and neutrons. Electrons are particles that are part of the constitution of the atom.
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g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days
Answer:
6.88 mg
Explanation:
Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄
The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.
175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P
Step 2: Calculate the rate constant for the decay of ³²P
The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.
k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹
Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days
For first-order kinetics, we will use the following expression.
ln P = ln P₀ - k × t
ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d
P = 6.88 mg
Sound waves travel the same speed through all mediums (solids, liqiuds, and gases).
A
True
B
False
assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the initial mixture at equilibrium? If not, in what direction must the reactionproceed to reach equilibrium? (Hint: You will need to use the value of Kc you determined in the lab
Answer:
The mixture is not in equilibrium, the reaction will shift to the left.
Explanation:
Based on the equilibrium:
Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺
kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]
Where [] are concentrations at equilibrium. The reaction is in equilibrium when the ratio of concentrations = kc
Q is the same expression than kc but with [] that are not in equilibrium
Replacing:
Q = [10.0M] [1.0M] / [0.1M] [0.1M]
Q = 1000
As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc
In an experiment, you added a base, NaOH, one mL at a time to 50 mL acetate buffer and recorded the pH. For the first 6 mL NaOH the pH increased from 4.5 to 4.9. At the 7th mL the pH was 6.6 and by the 8th mL the pH was 10.7. Knowing what you do about titrating acetate buffer with acid, is this experimental result what you expected or is it not expected
Answer:
yes the experimental result is the expected result .
Explanation:
When Titrating acetate buffer with acid the PH will decrease gradually from a more neutral PH to a more acidic level and this is because buffer solutions are prepared with weak acids and its conjugate base.
The results gotten from the continuous addition of base NaOH to the acetate buffer is the expected result because the base is been absorbed by the buffer solution and it is converted to a conjugate base of the buffer solution which will gradually increase the PH level of the solution as more conjugate base is formed due to the addition of more NaOH.
At 1630 hours you started an IV with 500 cc of D5W running at 60 gtt/min using microdrip
tubing (60 gtt/ce). You now receive an BAXTER IV Pump.
A)How many cc/hr should you set the pump for to keep the IV going at the same rate?
Answer:._ml/h
1. B) What time will the infusion be completed?
ions
Answer:
Answer:
0050 hours
Explanation:
The size of the drops is regulated by types of tubing in microdrip tubing 1cc(1ml ) forms 60 gtt.
Gtt in 500 cc D5W is 500×60 gtt.= 30000 gtt.
Time required to infuse whole fluid with the rate of 60gtt /min = 30000gtt÷60gtt/min = 500minute.
= 8hours 20 minute.
Fluid infusion started at 1630 hours after and it ends after 8 hours 30 minutes .
Therefore, infusion will be completed at 1630 hours +0830 hours.= 0050 hours.
if 7.90 mol of C5H12 reacts with excess O2, how many moles of CO2 will be produced by the following combustion reaction?
Answer:
If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.
Explanation:
The balanced reaction is:
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₅H₁₂: 1 moles O₂: 8 molesCO₂: 5 moles H₂O: 6 molesThen you can apply the following rule of three: if by stoichiometry 1 mole of C₅H₁₂ produces 5 moles of CO₂, then 7.9 moles of C₅H₁₂ will produce how many moles of CO₂?
[tex]amount of moles of CO_{2} =\frac{7.9 moles of C_{5}H_{12}*5 moles of CO_{2} }{1 mole of C_{5}H_{12} }[/tex]
amount of moles of CO₂= 39.5 moles
If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.
What is the difference between a physical change and a chemical change. Give an example of each.
Answer:
A physical change is a change in form.
A Chemical change is a change in materials.
Explanation:
Example of a physical change would be an ice cube meting.
Example of a chemical change would be mixing food coloring into a cup of water.
a. Compound A and compound B are constitutional isomers with molecular formula C3H7Cl. When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination rection predominates. Propose structures A and B.
b. An unknown compound with molecular formula C6H13Cl is treated with sodium ethoxide to produce 2,3-dimethyl-2-butene as the major product. Identify the structure of the unknown compound.
Answer:
história phkfk
Explanation:
guiooupigjdytrss
greater than 6
Less than or equal to 3
Odd
Not 0
3or9
I think it's 9 ................'
During a reaction in an aqueous solution, the concentration of bactants
decreases and the amount of products increases. How do these changes in
concentration affect the reaction rate?
A. The reaction rate decreases.
B. The reaction rate varies unpredictably.
C. The reaction rate increases.
D. The reaction rate stays the same.
Answer:
my define it will be turst me is c
The "nitrogen rule" of mass spectrometry requires a compound containing an odd number of nitrogens to have an odd-mass molecular ion and a compound containing an even number of nitrogens to have an even-mass molecular ion. What is the molecular formula of the CHN-containing compound pyrazine, M+ = 80? (The order of atoms should be carbon, then hydrogen, then others in alphabetical order.)
Answer:
C₄H₄N₂
Explanation:
Given that:
M+ = 80.
It implies that the number of nitrogen present in the molecule must also be even according to the Nitrogen rule.
So from the Formula CHN, the nitrogen will have to be 2 because if we make use of 4, it will exceed the given M+ which is 80.
∴
C₄ = 4 × 12 = 48
H₄ = 4 × 1 = 4
N₂ = 2 × 14 = 28
80
As such, the molecular formula of the compound is C₄H₄N₂
How are elements with similar properties grouped in the periodic table?
A. In the same half
B. In the same column
C. In the same row
D. In the same box
AAnswer:A
Explanation:
draw the structure of two acyclic compounds with 3 or more carbons which exhibits one singlet in the 1H-NMR spectrum
Answer:
attached below
Explanation:
Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum
a) Acetone CH₃COCH₃
Attached below is the structure
b) But-2-yne (CH₃C)₂
Attached below is the structure
A 2.9 kg model rocket accelerates at 15.3 m/s2 with a force of 44 N. Before launch, the model rocket was not moving. After the solid rocket engine ignited, hot gases were pushed out from the rocket engine nozzle and propelled the rocket toward the sky.
Which of Newton’s laws apply in this example?
Answer:
Newton's first and third law of Motion
Explanation:
The laws applying in the example Newton's first and third laws of Motion.
The first law states that any object at rest (ie. not moving) will stay at rest until it is forced to move by an external force. In this case, said force were the propulsion gases ignited.As the hot gases were pushed out from the engine nozzle, there was another force equal in magnitud but opposite in direction (as the gases went down, that force went upwards), said force is directly responsible for the rocket taking off. That is an example of the third law.Answer:
It Newtons first, second, and third laws
Explanation:
Como es la
fórmula química del agua
Answer:
h2o
Explanation:
Cu20(s) + C(s) - 2Cu(s) + CO(g)
To perform this synthesis, the team added 114.2 grams of Cu20 to 11.1 grams of C to form 87.1 grams of Cu.
In this copper synthesis reaction, what is the limiting reagent and the excess reagent?
Answer:
That means Cu2O is limiting reagent and C is excess reagent
Explanation:
Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.
To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.
Moles Cu2O -Molar mass: 143.09 g/mol-
114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O
Moles C -Molar mass: 12.01g/mol-
11.1g C * (1mol / 12.01g) = 0.924 moles C
That means Cu2O is limiting reagent and C is excess reagent
Although Joule is the unit of energy we normally use in science, people also use the Kilocalorie and the gram calorie when describing energy; especially where food energy is concerned. These are potentially confusing because 1 kilocalorie does not equal 1 gram calorie. Here's how it works:
• It takes 4.184 J to heat one gram of water 1 K
• It takes one gram calorie to heat one gram of water 1 K
• This means that 1 gram calorie = 4184
• 1 kilocaloric equals 1000 gram calories
• This means that 1 kilocalorie - 4184
The kilocalorie is also called the Dietary Calorie or Food Calorie. Convert your results to kilocalories per gram.
Answer:
• 1 kilocaloric equals 1000 gram calories
• 1 kilocalorie is equals to 4184 Joules of energy.
Explanation:
One kilocalorie is equals to 1000 gram calories because one kilo is equals to or have 1000 grams. One kilocalorie is equals to 4184 Joules of energy while on the other hand, one calorie is equals to 4.184 Joules of energy because one calorie is 1000 times smaller than Kilocalorie so calorie has also 1000 times lower energy than kilocalorie. Kilo is the prefix which means 1000 so we can say that One kilocalorie is equals to 1000 gram calories.
A mixture of coarse sand and sugar is 45.0 percent sand by mass. 120.0 grams (g) of the mixture is placed in a fine-mesh cloth bag and dunked repeatedly in Lake Michigan. After drying, the mass of the contents of the bag equals: ________.
A. 66.0 g
B. 120.0 g
C. 65.0 g
D. 72.00 g
E. 54.0 g
Answer:
Option E
Explanation:
From the question we are told that:
Amount of sand in percentage [tex]s_p=45%[/tex]
Sample size[tex]n=120g[/tex]
Note:After being dumped in the river repeatedly the sugar melts away leaving behind the insoluble sand
Generally the equation for Amount of sand content is mathematically given by
[tex]X=n*s_p[/tex]
[tex]X=120*\frac{45}{100}[/tex]
[tex]X=54g[/tex]
Therefore
After drying, the mass of the contents of the bag equals
[tex]X=54g[/tex]
Option E
Use the Conductivity interactive to identify each aqueous solution as a strong electrolyte, weak electrolyte, or nonelectrolyte. You are currently in a sorting module. Turn off browse mode or quick nav, Tab to items, Space or Enter to pick up, Tab to move, Space or Enter to drop.
Strong electrolyte Weak electrolyte Nonelectrolyte
NH3 NaCl HCI NaOH C12H22O
Explanation:
strong electrolyte- Nacl HCL NAOH
weak electrolyte- c12H22O, NH3
NaCl,HCl and NaOH are strong electrolytes while ammonia is a weak electrolyte and sucrose is a non-electrolyte.
What are electrolytes?It is a solution which consists of ions which are electrically conducting as a result of movement of ions.Class of electrolytes include most soluble salts,acids and bases which are dissolved in a polar solvent.On dissolution, they separate into the constituent ions.
There are 3 classes according to the nature of substance which results upon dissolution:
1) Strong electrolytes- Substances which on dissolution in a medium dissociate completely are strong electrolytes. eg: NaCl,HCl
2) Weak electrolytes- Substances which on dissolution in a medium dissociate partially are weak electrolytes. eg: NH₃
3)Non-electrolytes- Substances which do not dissociate on dissolution are non-electrolytes. eg: sucrose
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Which best expresses the uncertainty of the measurement 32.23 cm?
A.) ±0.05 cm
B.) 0.1 cm
C.) 1%
D.) ±0.01 cm?
Answer:
D.) ±0.01 cm?
Explanation:
Since 32.23 cm has two decimal places, the uncertainty is taken as one-half the last decimal pace.
The last decimal place is 0.03. Half of this is 0.03 cm/2 = 0.015 cm.
Since we cannot go below two decimal places, we ignore the 5 in 0.015 cm.
So, we have our uncertainty as 0.01 cm.
So, the best expression of the uncertainty in the measurement 32.23 cm is ± 0.01 cm.
So, the answer is D. which is ± 0.01 cm.
A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal in gm which combine with one gram of oxygen in each case
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
The cell potential of the following electrochemical cell depends on the pH of the solution in the anode half-cell:Pt(s)|H2(g, 1atm)|H+(aq, ?M)||Cu2+(aq,1.0M)|Cu(s)What is the pH of the solution if Ecell = 355 mV?
Answer:
0.51
Explanation:
Given the Nernst equation;
E= E° - 0.0592/n logQ
E= 355 mV or 0.355 V
E° = 0.34 - 0= 0.34 V
n= 2(two electrons were transferred in the process)
Equation of the reaction;
H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)
Substituting values;
0.355 = 0.34 - 0.0592/2 log([H^+]/1)
0.355 - 0.34 = - 0.0296 log [H^+]
0.015/-0.0296 = log [H^+]
Antilog (-0.5068) = [H^+]
[H^+] = 0.311 M
pH = -log[H^+]
pH= - log(0.311 M)
pH = 0.51
The potential difference between the half cell of the electrochemical cell is called cell potential. The pH of the solution at 355 mV will be 0.51.
What is an electrochemical cell?An electrochemical cell generates electricity from the redox chemical reactions occurring inside the cell.
The balanced chemical reaction is shown as,
[tex]\rm H_{2}(g) + Cu^{2+}(aq) \rightarrow 2H^{+}(aq) + Cu(s)[/tex]
Using the Nernst equation:
[tex]\rm E= E^{\circ} - \dfrac{0.0592}{n }logQ[/tex]
Given,
E = 0.355 V
E° = 0.34 V
n = 2
Substituting values in the above equation:
[tex]\begin{aligned} 0.355 &= 0.34 - \dfrac{0.0592}{2} \;\rm log(\dfrac{[H^{+}]}{1})\\\\0.355 - 0.34 &= - 0.0296 \rm \; log [H^{+}]\\\\\dfrac{0.015}{-0.0296} &= \rm \; log [H^{+}]\end{aligned}[/tex]
Solving further,
[tex]\begin{aligned} \rm Antilog (-0.5068)& = \rm [H^{+}]\\\\\rm [H^{+}] &= 0.311 \;\rm M \end{aligned}[/tex]
The pH of the solution is calculated as:
[tex]\begin{aligned} \rm pH &= \rm -log[H^{+}]\\\\&= \rm - log(0.311\; M)\\\\&= 0.51\end{aligned}[/tex]
Therefore, 0.51 is the pH of the solution.
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During a chemical reaction, an iron atom became the ion Fe2+. What happened to the iron atom?
Explanation:
Iron atom is been oxidised as it losses 2 electron to form 2 + ion.
Compare the solubility of silver chloride in each of the following aqueous solutions:
a. 0.10 M AgNO3 More soluble than in pure water.
b. 0.10 M NaCI Similar solubility as in pure water
c. 0.10 M KNO3 Less soluble than in pure water.
d. 0.10 M NH4CH3COO
Answer:
Compare the solubility of silver chloride in each of the following aqueous solutions:
a. 0.10 M AgNO3 More soluble than in pure water.
b. 0.10 M NaCI Similar solubility as in pure water
c. 0.10 M KNO3 Less soluble than in pure water.
d. 0.10 M NH4CH3COO
Explanation:
This is based on common ion effect.
According to common ion effect, the solubility of a sparingly soluble salt decreases in a solution containing common ion to it.
The solubility of AgCl(s) is shown below:
[tex]AgCl(s) <=> Ag^{+}(aq)+Cl^-(aq)[/tex]
So, when it is placed in:
a. 0.10 M AgNO3
Due to common ion effect Ag+, its solubility is less in this solution than in pure water.
b. 0.10 M NaCI :
Due to common ion effect Cl-, its solubility is less in this solution than in pure water.
c. 0.10 M KNO3 :
In this solution there is no presence of common ion.
So, the solubility of AgCl in this solution is similar to that of pure water.
d. 0.10 M NH4CH3COO:
In this solution, AgCl forms a precipitate.
So, the solubility of AgCl is more in this solution compared to pure water.
What would happen to the pressure of a closed sample of gas whose temperature increased while its volume decreased? Explain your reasoning in terms of the kinetic molecular theory of gases.
Answer:
As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase.
I hope this will help you if not soo sorry :)
how many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?
equation is H2 + Cl2 = 2HCl
Answer:
[tex]m_{HCl}=36.1gHCl[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:
[tex]n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2[/tex]
Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:
[tex]m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl[/tex]
Regards!
Aluminum reacts with excess copper(II) sulfate according to the unbalanced reaction
Al(s) + CuSO4(aq) −→
Al2(SO4)3(aq) + Cu(s)
If 2.98 g of Al react and the percent yield of
Cu is 46.4%, what mass of Cu is produced?
Answer in units of g.
Answer: The mass of Cu produced is 4.88 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of aluminum = 2.98 g
Molar mass of aluminum = 27 g/mol
Plugging values in equation 1:
[tex]\text{Moles of aluminum}=\frac{2.98g}{27g/mol}=0.1104 mol[/tex]
The given chemical equation follows:
[tex]2Al(s)+3CuSO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3Cu(s)[/tex]
By the stoichiometry of the reaction:
If 2 moles of aluminum produces 3 moles of Cu
So, 0.1104 moles aluminium will produce = [tex]\frac{3}{2}\times 0.1104=0.1656mol[/tex] of Cu
Molar mass of Cu = 63.5 g/mol
Plugging values in equation 1:
[tex]\text{Mass of Cu}=(0.1656mol\times 63.5g/mol)=10.516g[/tex]
The percent yield of a reaction is calculated by using an equation:
[tex]\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100[/tex] ......(2)
Given values:
% yield of product = 46.4 %
Theoretical value of the product = 10.516 g
Plugging values in equation 2, we get:
[tex]46.4=\frac{\text{Actual value of Cu}}{10.516g}\times 100\\\\\text{Actual value of Cu}=\frac{46.4\times 10.516}{100}\\\\\text{Actual value of Cu}=4.88g[/tex]
Hence, the mass of Cu produced is 4.88 g
.................. are microorganism used to improve soil fertility.
plz ams it correct
Answer:
Some bacteria like rhizobium and blue green algae are able to fix nitrogen gas from the atmosphere to enrich the soil with nitrogen compounds and increase its fertility. The nitrogen-fixing bacteria and blue green algae are called biological nitrogen fixers.
Answer:
Diazotrophic bacteria or Cyanobacteria
When metal X is treated with sodium hydroxide, a white precipitate A is obtained which is soluble in excess NaOH to give a soluble complex B. Compound A is soluble in dilute HCl to form compound C. When the compound A is heated strongly it gives compound D which is used to extract metal. a) Identify X, A, B, C, D supporting your answer(s) with appropriate chemical reactions. b) At which group and period does X fall?
Answer:
See explanation
Explanation:
If we look at the question closely, we will notice that the metal in question must be aluminum.
When aluminum is treated with sodium hydroxide, a precipitate, aluminium hydroxide is formed as follows;
Al(s) + 3NaOH(aq) ---> Al(OH)3(s) + 3Na(s)
In excess sodium hydroxide, the precipitate dissolves as follows;
Al(OH)3(s) + NaOH(aq) ----> [NaAlOH4]^-(aq)
The complex formed is sodium aluminum tetrahydroxo aluminate III.
The reaction of aluminum faith dilute hydrochloric acid occurs as follows to yield aluminum chloride;
2Al(s) + 6HCl(aq) ----> 2AlCl3(aq) + 3H2(g)
When aluminum metal is heated strongly, it yields aluminum oxide;
2Al(s) + 3O2(g) ---> Al2O3(s)