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Rufina [12.5K]
3 years ago
8

If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by ____

Chemistry
1 answer:
liraira [26]3 years ago
3 0

If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by two. It follows Charles law where in for a mixed gas of mass, the volume is directly proportional to the temperature at constant pressure.

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Product testing claims are called?
Tema [17]

Explanation:

They are called consumer testing or comparative testing, is a process of measuring the properties or performance of products. ... Product testing might be accomplished by a manufacturer, an independent laboratory, a government agency, etc. Often an existing formal test method is used as a basis for testing

hope this helps

7 0
3 years ago
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222 g of MTBE (CO(CH3)4) are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). Assume that the de
Eva8 [605]

Answer:

Explanation:

From the information given:

(a)

Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}

Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}

Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L

(b)

number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}

(c)

w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%

where; \\ \\ mass \  of \  (RFG) = 2L \times 0.70 g/mL \\ \\ mass \  of \  (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g

∴

w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}

(d)

Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}

Volume \ of \ MTBE = 300 \ mL\\

∴

v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}

(e)

From \the  \ given \  information; \\ \\ 2.5184 \ moles\ of  \ MTBE contain  \ 2.5184  \ mole of oxygen

∴

mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g

∴

\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%

\% w/w == 2.88\%

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3 years ago
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Give one example of a question that science would not be able to test. Then, explain how it could be changed into a testable que
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Testable = specific question that can be answered by hands on investigation.
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Which of the following is TRUE?
stiks02 [169]

Answer:

Explanation:

For a general equilibrium

aA +bB ⇔ cC + dD ,

the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.

Our reasoning here should be based on the fact that  Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).

So with this in mind, lets answer this question.

1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.

2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.

3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.

4. True: From our previous reasongs this is the true one.

5. False: If K is small, the equilibrium lies on the reactants side.

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