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3 years ago

What masses of KCl and H2O are needed to make 250 g of 5.00 % solution?

Chemistry

1 answer:

oksano4ka [1.4K]3 years ago

7 0

Answer:

w/w means weight/weight

Therefore, in a 250 g of solution, 5% needs to be KCl,

i.e. 5% x 250 = 12.5 g of KCl is needed.

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Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) <em>Alice heats the gas at constant volume until its pressure is doubled</em>:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) <em>Bob further heats the gas at constant pressure until its volume is doubled</em>:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

6 0

3 years ago

Methane gas and oxygen gas react to form water vapor and carbon dioxide gas. What volume of carbon dioxide would be produced by

Bas_tet [7]

Answer:

$V_{CO_2}=9.23m^3$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Now, as the stoichiometrical factors are in terms of mole but no information about neither the temperature nor the pressure is given, by means of the Avogadro's law, one could perform the stoichiometric calculations with the given volume as both the pressure and temperature remain the same, that is:

$V_{CO_2}=9.23m^3CH_4*\frac{1m^3CO_2}{1m^3CH_4} \\\\V_{CO_2}=9.23m^3$

Such 1:1 volume relationship equals the 1:1 molar relationship given in the chemical reaction in terms of their stoichiometric coefficients, therefore, the yielded volume of carbon dioxide is also 9.23m³

Best regards.

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4 years ago

What is the result of multiplying 2.5 times 10 to the power 10 by 3.5×10 to the power of -7

AURORKA [14]

Answer:

8.75 × 10³

Explanation:

Multiply the coefficients and the exponential terms separately; then multiply them together

2.5 × 3.5 = 8.75.

When you multiply exponential terms, you add the exponents.

10¹⁰ × 10⁻⁷ = 10⁽¹⁰⁻⁷⁾ = 10³

The product is 8.75 × 10³.

A scientific calculator does the operation automatically. The buttons may have different labels on your calculator, but the key sequence is usually something like

2.5 Exp 10 × 8.5 Exp +/- 7 = 8.75×10⁻³

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3 years ago

Which of these scales rates earthquake damage at a particular location? A. Focus B.Modified Mercalli C. Richter D. Moment magnit

allochka39001 [22]

The answer is C. Richter
the Richter scale is used to determine the magnitude of any earthworks

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3 years ago

An amide may be produced by reacting an acid chloride with

Elanso [62]

Answer:

An amide may be produced by reacting an acid chloride with ammonia.