Determine the number of atoms in 51.0 grams of sodium, Na. (The mass of one mole of sodium is 22.99 g.)
Answer:
The answer is
1.340 × 10²⁴ sodium atomsTo find the number of atoms of sodium we use the formula
N = n × L
where
n is the number of moles
N is the number of entities
L is Avogadro's constant which is
[tex]6.02 \times {10}^{23} [/tex]
We need to find the number of moles first
The formula is
[tex]n = \frac{m}{M} [/tex]
where
M is the molar mass
m is the mass
n is the number of moles
From the question
M = 22.9 g/mol
m = 51.0 g
[tex]n = \frac{51}{22.9} [/tex]
n = 2.227 moles
So the number of sodium atoms is
[tex]N = 2.227 \times 6.02 \times {10}^{23} [/tex]
We have the final answer as
1.340 × 10²⁴ sodium atomsHope this helps you
At an elevation where the boiling point of water is 93°C, 1.00 kg of water at 30°C absorbs 290.0 kJ from a mountain climber’s stove. Is this amount of thermal energy sufficient to heat the water to its boiling point? [cp of water = 4.18 J/(g · °C)] need more information to calculate can not be calculated even with more information no yes
Answer:
Yes, it will be enough.
Explanation:
We can calculate the heat (Q) required to heat 1.00 kg of water from 30°C to 93°C using the following expression.
Q = cp × m × ΔT
where,
cp: specific heat capacity of waterm: mass of waterΔT: change in the temperatureQ = cp × m × ΔT
Q = 4.18 J/g°C × 1.00 × 10³ g × (93°C-30°C)
Q = 263 kJ
Since 263 kJ are necessary, 290.0 kJ will be enough to heat the water.
The energy is sufficient to raise the temperature of the water to its boiling point.
We have the following information from the question;
Boiling point of water = 93°C
Mass of water = 1.00 kg or 1000 g
Heat capacity of water = 4.18 J/g · °C
Heat absorbed by water = 290.0 kJ or 290000 J
Initial temperature of the water = 30°C
Using the formula;
ΔH = mcθ
ΔH = Heat absorbed by the water
m = mass of the water
c = heat capacity of the water
θ = temperature rise (T2 - T1)
Substituting values;
290000 J = 1000 g × 4.18 J/g · °C (T2 - 30°C)
290000 = 4180T2 - 125400
T2 = 290000 + 125400/4180
T2 = 99.3°C
The energy is sufficient to raise the temperature of the water to its boiling point.
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how many moles of MgO are produced when .250 mol of Mg reacts completely with O2
Answer:
0.250 mol
Explanation:
The reaction between Mg and O2 is given by;
2Mg + O2 --> 2MgO
From the equation above; 2 moles of Mg reacts to form 2 moles of MgO.
0.250 mol of Mg would produce x mol of MgO.
2 = 2
0.250 = x
x = 0.250 * 2/2 = 0.250 mol
If a container were to have 24 molecules of C5H12 and 24 molecules of O2 initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion
Answer:
81 molecules
Explanation:
The reaction between C5H12 and O2 is a combustion reaction and is represented by the following equation;
C5H12 + 8O2 --> 5CO2 + 6H2O
The ratio of C5H12 to O2 from the above equation is 1 : 8.
Aplying the conditins of the question; 24 molecules each of C5H12 and O2 we have;
3C5H12 + 24O2 --> 15CO2 + 18H2O
This means we have 24 - 3 = 21 molecules of C5H12 that are unreacted.
Total molecules is given as;
3(C5H12) + 24(O2) + 15(CO2) + 18(H2O) + 21(Unreacted C5H12) = 81 molecules
Romans used calcium oxide, CaO, to produce a strong mortar to build stone structures. Calcium oxide was mixed wit ch reacted slowly with CO2 in the air to give CaCO3.
Ca(OH)2(s) +CO2(g) → CaCO3(s)+H20(g)
Required:
a. Calculate the standard enthalpy change for this reaction.
b. How much energy is evolved or absorbed as heat if 7.50 kg of Ca(OH)2 reacts with a stoichiometric amount of CO2.
Answer:
The given reaction is:
Ca(OH)₂ (s) + CO₂ (g) ⇒ CaCO₃ (s) + H₂O (g)
The ΔH°f of Ca(OH)₂ (s) is -986.09 kJ/mole, the ΔH°f of CO₂ (g) is -393.509 kJ/mol, the ΔH°f of CaCO₃ (s) is -1207.6 kJ/mol, and the ΔH°f of H₂O (g) is -241.83 kJ/mol.
ΔH°rxn = 1 × ΔH°f of CaCO₃ (s) + 1 × ΔH°f of H₂O (g) - 1 × ΔH°f of Ca(OH)₂ (s) - 1 × ΔH°f of CO₂ (g)
ΔH°rxn = 1 (-1207.6) + 1(-241.83) - 1 (-986.09) - 1 (-393.509)
ΔH°rxn = -69.831 kJ
b) The molecular mass of calcium hydroxide is 74.096 gram per mole.
The mass of calcium hydroxide given is 7.50 Kg or 7500 grams.
The number of moles of calcium hydroxide is,
n = Mass of Ca(OH)₂ / Molecular mass of Ca(OH)₂
n = 7500 / 74.1
n = 101.21 moles
As ΔH is negative, therefore, release of heat is taking place. Thus, when one mole of calcium hydroxide reacts, the heat released is -69.831 kJ. Therefore, 101.21 moles of calcium hydroxide will release the heat,
= 101.21 × 69.831 kJ
= 7.067 × 10³ kJ
2. You deposit the 500 ul from #1 into a solution with a final volume of 1200 uL. What is the final concentration of NaCl in molar? In molar?
Answer:
[tex]C_2=1.25 M[/tex]
Explanation:
Hello,
In this case, since the concentration in #1 is 3M, during a dilution process, the moles of the solute (NaCl) remains the same, just the concentration and volume change as shown below:
[tex]n_1=n_2\\\\C_1V_1=C_2V_2[/tex]
In such a way, as the final volume is 1200 microliters, the resulting concentration turns out:
[tex]C_2=\frac{C_1V_1}{V_2}=\frac{3M*500\mu L}{1200\mu L}\\ \\C_2=1.25 M[/tex]
Best regards.
What are the signs of the enthalpy change (ΔH°) and the entropy change (ΔS°) for the condensation of CS2(g)?
Answer:
∆H is negative
∆S is negative
Explanation:
The condensation of CS2 implies a phase change from gaseous state to liquid state. The energy of the gaseous particles is greater than that of the liquid particles hence energy is given out when a substance changes from gaseous state to liquid state hence the process is exothermic and ∆H is negative.
Changing from gaseous state to liquid states leads to a decrease in entropy hence ∆S is negative. Liquid particles are more orderly than particles of a gas.
Consider the follow scenario of 15.3 g of NaCl was dissolved in 155.0 g of water.
1. What is the total mass of the solution?
2. What fraction of the total is Naci?
3. What percent of the total is Naci?
4. Use your percent to determine how many grams of NaCl are contained in 100 g of solution.
5. Determine how many grams of NaCl are in 38.2 g of the solution described at the top of model 1.
6. Use the appropriate two conversion factors to find what volume of this solution you would need to have exactly 2.00g NaCl. The density is 1.07g/mL.
Answer:
Total mass: 170.3 g
Fraction of NaCl: 0.089%
Percent of NaCl: 8.98%
3.43 g of NaCl in 38.2 g of solution
1 mL . (170.3 g of solution / 1.07 g solution) = 159.1 mL
159.1 mL . (2 g NaCl / 15.3 g NaCl) = 20.8 mL
Explanation:
Our scenario is 15.3 g of NaCl in 155 g of water
Total mass: 15.3 g + 155 g = 170.3 g of solution
Our solute is NaCl - Our solvent is water.
To determine the fraction we divide:
15.3 g / 170.3 g = 0.0898
To determine percent, we multiply the fraction by 100
0.089 . 100 = 8.98 %
We can make a conversion factor to determine the mass of NaCl in 38.2 g of solution. If 15.3 g of NaCl are in 170.3 g of solution and we need 38.2 g, we can propose → (15.3 / 170.3) . 38.2 = 3.43 g of NaCl
The conversion factors are to find what volume of solution is on 2g of NaCl are:
Density data always reffers to solution. So 1.07 grams of solution are contained in 1 mL of solution
1 mL . 170.3 g of solution / 1.07 g solution = 159.1 mL
This is the volume for our 15.3 g of NaCl so:
159.1 mL . (2 g NaCl / 15.3 g NaCl) = 20.8 mL
How would you make a 30% ethanol solution?
Answer:
You need 30 ml of alcohol and 70 ml of water.
Explanation:
You want to end up with 100ml of liquid, 30% of which is alcohol. 30% of 100ml is 30/100 100 = 30 ml.
A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?
Answer:
Normality N = 0.2 N
Explanation:
Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.
Normality is represented by N.
Mathematically, we have :
[tex]\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}[/tex]
Given that:
number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent
volume of solution (HCl) = 450 mL 450 × 10⁻³ L
[tex]\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}[/tex]
Normality N = 0.2 N
What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose?
Answer:
[tex]T_f=-2.58\°C[/tex]
Explanation:
Hello,
In this case, we can compute the the freezing point depression by using the following formula:
[tex]T_f-T_0=-i*m*Kf[/tex]
Whereas the freezing point of pure water is 0 °C van't Hoff factor for glucose is 1, the molality is computed as shown below and the freezing point constant of water is 1.86 °C/m:
[tex]m=\frac{25.0g\ glucose*\frac{1mol\ glucose}{180g\ glucose} }{100g*\frac{1kg}{1000g} }\\ \\m=1.39m[/tex]
Thus, the freezing point of the solution is:
[tex]T_f=T_0-i*m*Kf\\\\T_f=0\°C-1*1.39m*1.86\frac{\°C}{m}\\ \\T_f=-2.58\°C[/tex]
Regards.
how are sound wave different from the waves in the sea or the ripples on water mark brainliest
Answer:
Sound wave are not solid nor liquid nor gas it is invisible and goes nearly through any thing.
While waters ripples are liquid which can be easily moved by anything
The 3d energy level in hydrogen has how many distinct states with different values of the quantum number m
The question is incomplete, the complete question is;
The 3d energy level in hydrogen has how many distinct states with different values of the quantum number m?
A. 3
B. 5
C. 6
D. 2
E. 4
Answer:
B. 5
Explanation:
The magnetic quantum number is used in describing the actual orientation of orbitals in space. The name 'magnetic quantum number' was coined because it describes the effect of different orientations of orbitals which was initially observed in the presence of an external magnetic field.
The d orbital can exhibit five orientations corresponding to five values of the magnetic quantum number, these are; -2,-1,0,1,2 hence the answer above.
In the Lewis structure for ICl2–, how many lone pairs of electrons are around the central iodine atom?
a. 0
b. 1
c. 2
d. 3
e. 4
Answer:
where is rhe structure
Which path will a carbon atom most likely travel from CO2 in the atmosphere to glucose in the cell of a secondary consumer
Answer:
See the answer below
Explanation:
The carbon will have to travel in the form of CO2 from the atmosphere to a primary producer (green plant), from there to a primary consumer (herbivorous animal), and finally to a secondary consumer.
The primary producer (a green plant) would fix the carbon in the CO2 to carbohydrate through a process known as photosynthesis. The equation of the process is as shown below:
[tex]6 CO_2 + 6 H_2O --> C_6H_1_2O_6 + 6 O_2[/tex]
The carbon, now in the form of carbohydrate, would then be picked up by an animal (a primary consumer) that feeds on the green plant. The carbon would eventually get into a secondary consumer when the secondary consumer feeds on the primary consumer that fed on the green plant.
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:[tex]T_f=33.85\°C[/tex]
Explanation:
Hello,
In this case, we can write the following relationship, explaining that the lost by the hot water is gained by the cold water:
[tex]Q_{hot,W}=-Q_{cold,W}[/tex]
Which in terms of mass, specific heat and temperatures, we have:
[tex]m_{hot,W}Cp_{W}(T_f-T_{hot,W})=-m_{cold,W}Cp_{W}(T_f-T_{cold,W})[/tex]
Whereas the specific heat of water is cancelled out to obtain the following temperature, considering that the density of water is 1 kg/L:
[tex]T_f=\frac{m_{hot,W}T_{hot,W}+m_{cold,W}T_{cold,W}}{m_{hot,W}+m_{cold,W}}\\\\T_f=\frac{5.0kg*80\°C+60kg*30\°C}{5.0kg+60kg} \\\\T_f=33.85\°C[/tex]
Regards.
If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
Answer:
The correct answer would be : 33.8 g
Explanation:
Molar mass of ammonia,
Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)
= 1*14.01 + 3*1.008 = 17.034 g/mol
mass(NH3)= 25.0 g (given)
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(25.0 g)/(17.034 g/mol)
= 1.468 mol
Now,
Molar mass of O2
= 32 g/mol
mass(O2)= 45.0 g
similar as ammonia
n (O2)=(45.0 g)/(32 g/mol)
= 1.406 mol
Balanced chemical equation is:
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
1.83456 mol of O2 is required for 1.46765 mol of NH3
by the calculation we have only 1.40625 mol of O2
Thus, the limiting agent will be - O2
now the Molar mass of NO,
= 1*14.01 + 1*16.0
= 30.01 g/mol (similar formula used for NH3)
Balanced equation :
mol of NO formed = (4/5)* moles of O2
= (4/5)×1.40625 (from above calculation)
= 1.125 mol
mass of NO = number of moles × molar mass
= 1.125*30.01
= 33.8 g
Thus, the correct answer would be : 33.8 g
The amount of nitrogen oxide that can be formed in the given mass is 44.12 g.
The given parameters;
mass of ammonia, NH₃ = 25.0 gmass of oxygen, O₂ = 45.0 gThe reaction of the ammonia and oxygen is written as follows;
[tex]4NH_3(g) \ + \ 5O_2 (g) \ --> \ 4NO (g) \ + \ 6H_2O(g)\\\\[/tex]
Molar mass of NH₃ = (14) + (3 x 1) = 17 g/mol
Molar mass of NO = (14) + 16 = 30 g/mol
4(17 g/mol) of NH₃ ------------------ 4(30)
25 g/mol of NH₃ --------------------- ?
[tex]= \frac{4(30) \times 25}{4(17)} \\\\= 44.12 \ g[/tex]
Thus, the amount of nitrogen oxide that can be formed in the given mass is 44.12 g.
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4-methyl-3-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.
Answer:
Structure in attachment.
Explanation:
The oxymercuration-demercuration of an asymmetric alkene usually produces the Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two carbon atoms, but there is more positive charge on the more substituted carbon atom, where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.
In hydroboration, borane adds to the double bond in one step. Boron is added to the less hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.
The idea that the moon, sun, and known planets orbit Earth is called the _______________ model of the universe.
Answer:
spherical model of the universe
If 20.6 grams of ice at zero degrees Celsius completely change into liquid water at zero degrees Celsius, the enthalpy of phase change will be positive. TRUE FALSE
Answer:
TRUE.
Explanation:
Hello,
In this case, since the fusion enthalpy of ice is +333.9 J/g and the fusion entropy is defined as:
[tex]\Delta _{fus}S=\frac{m*\Delta _{fus}H}{T_{fus}}[/tex]
We can compute it considering the temperature (0 °C) in kelvins:
[tex]\Delta _{fus}S=\frac{20.6g*333.9J/g}{(0+273)K}\\\\\Delta _{fus}S=25.2J/K[/tex]
Therefore answer is TRUE.
Best regards.
Use the periodic table to determine the number of valence electrons in each of the following elements.
Na:
E:
V:
Ar:
Answer:
Na: 1
F: 7
V: 5
Ar: 8
C: 4
Explanation:
The number of valence electrons by using periodic table are Na has 1, F has 7, V has 5, Ar has 8 and C has 4 valence electron.
What is periodic table ?The chemical elements are arranged in rows and columns in the periodic table, sometimes referred to as the periodic table of the elements. It is frequently used in physics, chemistry, and other sciences, and is frequently regarded as a symbol of chemistry.
Because of the orderly arrangement of the elements, it is known as the periodic table. They're arranged in rows and columns, as you'll see. Periods and groups are the names given to the horizontal rows and the vertical columns, respectively.
A system for arranging the chemical elements is the periodic table. The fundamental components of all matter are the chemical elements. The atomic number is a distinct characteristic of each chemical element. This figure is based on how many protons there are in each of the element's atoms.
Thus, The number of valence electrons by using periodic table are Na has 1, F has 7, V has 5, Ar has 8 and C has 4 valence electron.
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Chlorine dioxide reacts in basic water to form chlorite and chlorate according to the following chemical equation:
2ClO2(aq) + 2OH−(aq) → ClO−2(aq) + ClO−3(aq) + H2O(l)
Under a certain set of conditions, the initial rate of disappearance of chlorine dioxide was determined to be 2.30 × 10−1 M/s. What is the initial rate of appearance of chlorite ion under those same conditions?
Answer: The initial rate of appearance of chlorite ion under those same conditions is [tex] 1.15\times 10^{-1}M/s[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]2ClO_2(aq)+2OH^-(aq)\rightarrow ClO_2^{-}(aq)+ClO_3^{-}(aq)+H_2O(l)[/tex]
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of =
Rate in terms of appearance of =
The rate of disappearance of chlorine dioxide = [tex]2.30\times 10^{-1} M/s[/tex]
[tex]\frac{d[ClO_2]}{2dt}=\frac{d[ClO_2^{-}]}{dt}[/tex]
[tex]\frac{2.30\times 10^{-1}}{2}=\frac{d[ClO_2^{-}]}{dt}[/tex]
[tex]\frac{d[ClO_2^{-}]}{dt}=1.15\times 10^{-1}M/s[/tex]
The initial rate of appearance of chlorite ion under those same conditions is [tex] 1.15\times 10^{-1}M/s[/tex]
If the initial rate of disappearance of ClO₂ is 2.30 × 10⁻¹ M/s, the rate of appearance of ClO₂⁻ is 1.15 × 10⁻¹ M/s.
Chlorine dioxide reacts in basic water to form chlorite and chlorate according to the following chemical equation:
2 ClO₂(aq) + 2 OH⁻(aq) → ClO₂⁻(aq) + ClO₃⁻(aq) + H₂O(l)
In this problem, we want to find an initial rate of reaction.
What is the rate of reaction?The rate of reaction is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time.
We can relate the rate of disappearance of ClO₂ and the rate of appearance of ClO₂⁻, using the molar ratios.
What are the molar ratios?Molar ratios state the proportions of reactants and products that are used and formed in a chemical reaction.
The molar ratio of ClO₂ to ClO₂⁻ is 2:1.
If the initial rate of disappearance of ClO₂ is 2.30 × 10⁻¹ M/s, the rate of appearance of ClO₂⁻ is:
2.30 × 10⁻¹ mol ClO₂/L.s × (1 mol ClO₂⁻/2 mol ClO₂) = 1.15 × 10⁻¹ M/s
If the initial rate of disappearance of ClO₂ is 2.30 × 10⁻¹ M/s, the rate of appearance of ClO₂⁻ is 1.15 × 10⁻¹ M/s.
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There are always attractive forces between a collection of atoms or molecules which may not be negligible as suggested by the kinetic molecular theory. The strength of these forces depend upon the nature of the atom or molecule. If a mole of gas is at STP and there are very strong attractive intermolecular forces between the gas particles, the volume will be _________.
Answer:
Negligible
Explanation:
According to the kinetic theory of gases, the degree of intermolecular interaction between gases is minimal and gas molecules tend to spread out and fill up the volume of the container.
If the attraction between gas molecules increases, then the volume of the gas decreases accordingly. This is because, gas molecules become highly attracted to each other.
This intermolecular attractive force may be so strong, such that the actual volume of the gas become negligible compared to the volume of the container.
When converting 3.45 pounds to grams you need to know that 1 pound is equal to 453.6 grams. What would go on the bottom (denominator) of the first conversion factor? *
Answer:
it would be 3.45lb/1 *454grams /lb
Explanation:
Put the following in order from least to most dense. Water, steam, salt water, ice
What is the atomic mass for Helium (He)? Question 5 options: 8 2 3 4
Answer:
the answer is D
Explanation:
Answer; 4
is the atomic mass
A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentration of I2 was 6.29×10^−4M. Calculate the Kc at 1000K for:
H2(g)+I2(g)⇌2HI(g)
Answer:
The answer is "29.081"
Explanation:
when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.
[tex]\text{calculating the initial HI}= \frac{mol}{V}[/tex]
[tex]=\frac{9.3 \times 10 ^ -3}{2}[/tex]
[tex]=0.00465 \ Mol[/tex]
[tex]\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }[/tex]
Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:
[tex]HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\[/tex]
It is defined that:
[tex]I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\[/tex]
[tex]HI \ eq= 0.00465 - 2x \\[/tex]
[tex]=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M[/tex]
Now, we calculate the position:
For the reaction [tex]H 2(g) + I 2(g)\rightleftharpoons 2HI(g)[/tex], you can calculate the value of Kc at 1000 K.
data expression for Kc
[tex]2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}[/tex]
[tex]= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386[/tex]
calculating the reverse reaction
[tex]H_2(g) + I_2(g)\rightleftharpoons 2HI(g)[/tex]
[tex]Kc = \frac{1}{Kc} \\\\[/tex]
[tex]= \frac{1}{0.034386}\\ \\= 29.081\\[/tex]
The Kc of the reaction is 40.
Molarity of the HI = 9.30×10^−3mol/ 2.00 L = 4.65 × 10^-3 M
Let the concentrations of I2 and H2 be x, but we are told in the question that 6.29×10^−4M was present at equilibrium.
The molarity of HI at equilibrium now becomes; 4.65 × 10^-3 M - 6.29×10^−4M
= 4 × 10^-3 M
But;
Kc = [HI]^2/[H2] [I2]
Kc = ( 4 × 10^-3)^2/(6.29×10^−4)^2
Kc = 40
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What is the activation energy for a reaction which proceeds 50 times as fast at 400 K as it does at 300 K? Answer in units of J/mol rxn."
Answer:
Activation energy for the reaction is 39029J/mol
Explanation:
Arrhenius equation is an useful equation that relates rate of reaction at two different temperatures as follows:
[tex]ln\frac{K_2}{K_1} = \frac{-Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Where K₁ and K₂ are rate of reaction, Ea is activation energy and R is gas constant (8.314J/molK
If the reaction at 400K is 50 times more faster than at 300K:
K₂/K₁ = 50 where T₂ = 400K and T₁ = 300K:
[tex]ln50 = \frac{-Ea}{8.314J/molK} (\frac{1}{400K} -\frac{1}{300K} )[/tex]
[tex]ln 50 = 1x10^{-4}Ea[/tex]
Ea = 39029 J/mol
Activation energy for the reaction is 39029J/mol
The activation energy for this chemical reaction is equal to 39,029.24 J/mol.
Given the following data:
Rate of reaction = 50Final temperature = 400 KInitial temperature = 300 KIdeal gas constant, R = 8.314 J/molK
To determine the activation energy for this chemical reaction, we would use the Arrhenius' equation:
Mathematically, Arrhenius' equation is given by the formula:
[tex]ln\frac{K_2}{K_1} = \frac{-E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})[/tex]
Where:
K is the rate of chemical reaction.[tex]E_a[/tex] is the activation energy.R is the ideal gas constant.T is the temperature.Substituting the given parameters into the formula, we have;
[tex]ln50 = \frac{-E_a}{8.314} (\frac{1}{400} - \frac{1}{300})\\\\3.9120 = \frac{-E_a}{8.314} (\frac{-1}{1200})\\\\3.9120 = \frac{E_a}{9976.8} \\\\E_a = 9976.8 \times 3.9120\\\\E_a = 39,029.24 \;J/mol[/tex]
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Which statement best describes the types and locations of particles that make up the atom? A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom. B. The neutral-charged neutrons, positive-charged protons, and negative-charged electrons are all found within the nucleus of the atom. C. The negative-charged electrons and positive-charged protons are found within the nucleus, and the neutral-charged neutrons orbit outside the nucleus of the atom. D. The negative-charged neutrons and positive-charged protons are found within the nucleus, and the neutral-charged electrons orbit outside the nucleus of the atom.
Answer:
A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom.
Explanation:
Before examining the options, it's important to know that the atom is made up of three sub atomic particles which are; Protons, Neutrons and Electrons.
The protons are positively charged and are found in the nucleus of an atom. The Neutron A=are neutral in terms of charge and are also found in the nucleus of an atom. The electrons on the other hand are negatively charged and are found outside the nucleus if an atom, more specifically on the orbitals.
The option that best describes this is; option A.
Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
A. 75.0 mL of 0.10 MHF; 55.0 mL of 0.15 MNaF
B. 150.0 mL of 0.10 MHF; 135.0 mL of 0.175 MHCl
C. 165.0 mL of 0.10 MHF; 135.0 mL of 0.050 MKOH
D. 125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 MCH3NH3Cl
E. 105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
A buffer is a solution that mitigates against changes in acidity/alkalinity.
A buffer consists of a weak acid and its conjugate base. Also, a buffer can be formed from a weak base and its conjugate acid. A buffer is a solution that helps to mitigate against changes in acidity and alkalinity.
Let us now examine the solution mixtures listed in the question:
75.0 mL of 0.10 MHF; 55.0 mL of 0.15 MNaF. This can work as a buffer solution because it contains a weak acid (HF) and its conjugate base(F^-).150.0 mL of 0.10 MHF; 135.0 mL of 0.175 MHCl will not function as a buffer solution 165.0 mL of 0.10 MHF; 135.0 mL of 0.050 MKOH will not function as a buffer solution125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 MCH3NH3Cl will not function as a buffer solution105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl will not function as a buffer solution.Learn more: https://brainly.com/question/13439771