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ANTONII [103]
2 years ago
5

Help??

Chemistry
1 answer:
kvasek [131]2 years ago
4 0

Answer:

Mass of sodium chloride decomposed = 24.54 g

Explanation:

Given data:

Mass of sodium chloride decomposed = ?

Mass of chlorine gas formed = 15 g

Solution:

Chemical equation:

2NaCl      →         2Na + Cl₂

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 15 g/ 71 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of Cl₂ with NaCl from balance chemical equation.

                    Cl₂            :              NaCl

                      1              :                2

                      0.21         :            2×0.21 = 0.42 mol

Mass of Sodium chloride decompose:

Mass = number of moles × molar mass

Mass = 0.42 mol × 58.44 g/mol

Mass = 24.54 g

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Metallic I’m pretty sure. :)
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the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh
Harman [31]

Answer:

\boxed{\text{10 L}}

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}

Calculation:

\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}

4 0
3 years ago
Helium gas diffuses 4 times as fast as an unknown gas. What is relative molecular mass of the gas​
Inessa05 [86]

The relative molecular mass of the gas​ : 64 g/mol

<h3>Further explanation</h3>

Given

Helium rate = 4x an unknown gas

Required

The relative molecular mass of the gas​

Solution

Graham's Law

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

r₁=4 x r₂

r₁ = Helium rate

r₂ = unknown gas rate

M₁= relative molecular mass of Helium = 4 g/mol

M₂ = relative molecular mass of the gas​

Input the value :

\tt \dfrac{4r_2}{r_2}=\sqrt{\dfrac{M_2}{4} }\\\\16=\dfrac{M_2}{4}\\\\M_2=64~g/mol

7 0
2 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
2 years ago
HNO3 and H2CO3 are examples of ?
Basile [38]
A) acids because they start with h
7 0
3 years ago
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