All categories

3 years ago

For the reaction represented by the equation 2h+O2– 2H2O, how may grams of water are produced from 6.00 mol of hydrogen?

Chemistry

1 answer:

Pepsi [2]3 years ago

3 0

Answer:

Mass = 108 g

Explanation:

Given data:

Mass of water produced = ?

Number of moles of hydrogen = 6.00 mol

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Now we will compare the moles of hydrogen with water.

H₂ : H₂O

2 : 2

6 : 6

Mass of water:

Mass = number of moles × molar mass

Mass = 6 mol × 18 g/mol

Mass = 108 g

You might be interested in

If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?

valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA H⁺ + A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid} H⁺(aq) + A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

6 0

3 years ago

What effect does nuclear radiation have on atoms

Diano4ka-milaya [45]

it is C it splits atoms

7 0

3 years ago

Read 2 more answers

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter?

NARA [144]

Answer:

ΔH = $q_{p}$

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

$q_{com}$ = $q_{p}$

where

$q_{p}$ = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

$q_{com}$ = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is $q_{p}$

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = $q_{p}$ + 0

==> ΔH = $q_{p}$

5 0

3 years ago

HELP PLEASE !!!!

love history [14]

Answers are:
-4 for C in CH4, because carbon has greater electronegativity than hydrogen and he attracts shared electrons.
+4 for C in CO2, because carbon has smaller electronegativity than oxygen.
+1 for H in both CH4 and H2O, because hydrogen has amaller electronegativity than both carbon and oxygen.

7 0

3 years ago

Read 2 more answers

The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react

Kazeer [188]

Answer:

The percent yield of chloro-ethane in the reaction is 82.98%.

Explanation:

$C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl$

Moles of ethane = $\frac{300.0 g}{30 g/mol}=10 mol$

Moles of chlorine gases = $\frac{650.0 g}{71 .0 g/mol}=9.1549 mol$

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

$\frac{1}{1}\times 9.1549 mol=9.1549 mol$ of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

$\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%$

The percent yield of chloro-ethane in the reaction is 82.98%.

6 0

3 years ago