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DENIUS [597]
2 years ago
14

How many grams of CO2 are produced if 2.09 moles of hydrochloric acid are reacted with

Chemistry
1 answer:
Anit [1.1K]2 years ago
6 0

Answer:

The molar mass of CO2 = 44.01 g/mol

Explanation:

Start with 2.09 moles of HCl and convert to moles of CO2 For every 2 moles of HCl used, 1 mole of CO2 is formed.

2.09 mol HCl •(1 mol CO2/2 mol HCl) = 1.045 mol CO2

Now convert moles of CO2 to grams using the molar mass.

1.045 mol CO2 • 44.01 g/mol = 45.99 grams CO2

3 significant figures = 46.0 grams CO2

Hope This Helped

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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
In a mixture of hydrogen and nitrogen gases, the mole fraction of nitrogen is 0.333. If the partial pressure of hydrogen in the
notka56 [123]

Answer:

P_T=112.4torr

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

P_{H_2}=x_{H_2}*P_T

And can be solved for the total pressure as follows:

P_T=\frac{P_{H_2}}{x_{H_2}}

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667

Then, we can plug in to obtain the total pressure:

P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr

Regards!

4 0
3 years ago
I need to find the orbit notation<br> 1. Ne - 1s 2s 2p 3s 3p 4s 3d 4p 5s​
weeeeeb [17]

Answer:

1s*2 2s*2 2p*6

Explanation:

atomic number of neon is 10 so the s-shell contains 2 electron and p-shell contains 6

3 0
3 years ago
The smallest ionic radius among the following. <br> Li<br> Mg<br> Na+<br> be+
dimaraw [331]
The correct answer is Be+

That is because it lost a single electron but still has the same number of protons, and thus the effective charge attracting each electron is greater, which in turn makes the radius even smaller
5 0
3 years ago
An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is
Juli2301 [7.4K]

Answer:

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}  

Explanation:

The gas ideal law is  

PV= nRT (equation 1)

Where:

P = pressure  

R = gas constant  

T = temperature  

n= moles of substance  

V = volume  

Working with equation 1 we can get  

n =\frac{PV}{RT}

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

\frac{m}{mw} =\frac{PV}{RT}  or  

m =\frac{P*V*mw}{R*T}   (equation 2)

The cylindrical container has a constant pressure p  

The volume is the volume of a cylinder this is

V =(pi)*r^{2}*h

Where:

r = radius  

h = height  

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is  V =(pi)*r^{2}*h

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:  

T =\frac{T_{1} + T_{O}}{2}  

Replacing these values in the equation 2 we get:

m =\frac{P*V*mw}{R*T}   (equation 2)

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}    

8 0
3 years ago
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