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Reptile [31]
3 years ago
14

Determine the mass of CO2 gas produced when 8.24 g of NaHCO3 is added to a solution that contains 4.41 g of HCl.

Chemistry
1 answer:
hjlf3 years ago
4 0

The mass of CO₂ gas = 4.312 g

<h3>Further explanation</h3>

Reaction

NaHCO₃ (s) + HCl (aq) → NaCl (aq) + H₂O (aq) + CO₂ (g)

Find limiting reactant

mol NaHCO₃

\tt \dfrac{8.24}{84,007 }=0.098

mol HCl

\tt \dfrac{4.41}{36.5}=0.121

Limiting reactant :  NaHCO₃ (smaller)

mol CO₂ = mol NaHCO₃ = 0.098

mass CO₂ :

\tt 0.098\times 44=4.312~g

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What volume would 3 moles of hydrogen gas occupy at stp?
Alchen [17]
22.4 L<span>So, if 1 mole occupies 22.4 L, the imediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L. In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas.</span>
3 0
3 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
Help plisssssssssss I need helppp
Makovka662 [10]

Answer:

Explanation:

you would have to look more around the page, for example look at some ways that you can right down.

6 0
3 years ago
Consider the following gas phase reaction:
snow_tiger [21]

Answer:

2NO(g) + O2(g) --> 2NO2(g)

now 400 ml of NO × 2 mol of NO2/2 mol of NO

= 400 ml of NO2

now 500 ml of O2 × 2 mol of NO2/1 mol of O2

= 1000 ml of NO2

now 400 ml of NO2 × 1 mol of O2/2 mol of NO

= 200 ml

subtract that from 500 ml of total i.e. 500-200 =300 ml

The total volume of the reaction mixture is 1000 ml -300ml = 700 ml

6 0
2 years ago
Balance the equations.<br> Zn+ HCl →<br> ZnCl2 +
motikmotik

Answer:

Zn + 2HCl → ZnCl2 + H2

Explanation:

Zn + HCl → ZnCl2 +

The complete equation is given below:

Zn+ HCl → ZnCl2 + H2

Now we can balance the equation by doing the following:

There are 2 atoms of Cl and 2 atoms of H on the left. This can be balanced by putting 2 in front of HCl as shown below:

Zn + 2HCl → ZnCl2 + H2

7 0
3 years ago
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