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3 years ago

Please answer this...

Chemistry

1 answer:

Licemer1 [7]3 years ago

3 0

Answer: 388.5 L CO2

777 L SO2 * 1 mol SO2 / 22.4 L SO2 * 1 mol CO2 / 2 mol SO2 * 22.4 L CO2 / 1 mol CO2 = 388.5 L CO2

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Suppose an egyptian mummy is discovered in which the amount of carbon-14 present is only about oneone-fifthfifth the amount fo

poizon [28]

We are given the following equation:

y = y0 e^-0.0001216 t

where y = 1/5 y0, y0 is the original amount

So solving for time t:

1/5 y0= y0 e^-0.0001216 t

t = 13,235.51 years

So the human died about 13235.5 years ago

3 0

3 years ago

what mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?

LuckyWell [14K]

Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.

7 0

3 years ago

How many moles of oxygen gas, o2, are in a storage tank with a volume of 1.000×105 l at stp?

Mumz [18]

STP means standard temperature and pressure which is equivalent to 273 K and 1 atm, respectively. Assuming ideal gas behavior, the solution for this problem is as follows:

PV = nRT
Solve for n,
n = RT/PV
n = (0.0821 L-atm/mol-K)(273 K)/(1 atm)(1×10⁵ L)
<em>n = 2.24×10⁻⁴ moles</em>

6 0

3 years ago

Read 2 more answers

A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch

ivanzaharov [21]

Answer:

The reactions free energy $\Delta G = -49.36 kJ$

Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

The pressure of (Cl) gas is $P_{Cl} = 9.15 \ atm$

The pressure of nitrosly chloride (NOCl) is $P_{(NOCl)} = 7.70 \ atm$

The reaction is

$2NO_{(g)} + Cl_2 (g)$ ⇆ $2 NOCl_{(g)}$

From the reaction we can mathematically evaluate the $\Delta G^o$ (Standard state free energy ) as

$\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}$

The Standard state free energy for NO is constant with a value

$\Delta G^o _{NO} = 86.55 kJ/mol$

The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

The Standard state free energy for $NOCl$ is constant with a value

$\Delta G^o _{NOCl} =66.1kJ/mol$

Now substituting this into the equation

$\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6$

$= -43 kJ/mol$

The pressure constant is evaluated as

$Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }$

Substituting values

$Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}$

$= 0.0765$

The free energy for this reaction is evaluated as

$\Delta G = \Delta G^o + RT ln Q$

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

T is temperature in K with a given value of $T = 25+273 = 298 K$

Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

$= -43-6.36$

$\Delta G = -49.36 kJ$

4 0

3 years ago

Using the molarity equation, calculate how many grams of salt you need in order to make 80ml of a 2M solution (MW = 58.44g/mole)

ollegr [7]

Answer:

9.35g

Explanation:

The molarity equation establishes that:

$\textrm{molarity}=\frac{\textrm{moles of solute}}{\textrm{liters of solution}}$

So, we have information about molarity (2M) and volume (80 ml=0.08 l), with that, we can find the moles of solute:

$\textrm{moles of solute}=\textrm{molarity}*\textrm{liters of solution}$

$\textrm{moles of solute}= 0.08 \textrm{ l} *2\textrm{ M} = 0.16 \textrm{ mol}$

The mathematical equation that establishes the relationship between molar weight, mass and moles is:

$\textrm{molar weight}= \frac{\textrm{mass}}{\textrm{moles}}$

$\textrm{MW}= \frac{\textrm{m}}{\textrm{n}}$

We have MW (58.44g/mole) and n (0.16 mol), and we need to find m (grams of salt needed) to solve the problem:

$\textrm{m} = \textrm{MW * n}= 58.44\frac{\textrm{g}}{\textrm{mol}} * 0.16 \textrm{ mol} = 9.35 \textrm{ g}$

8 0

4 years ago