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garri49 [273]
3 years ago
6

Which statement is true?

Chemistry
1 answer:
dimulka [17.4K]3 years ago
6 0

Answer:

A I think

Explanation:

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A scientist in London and a scientist in California both conducted independent investigations that produced similar results. Wha
VashaNatasha [74]
Then the data had to be the same.
6 0
3 years ago
Read 2 more answers
A mixture of Ar, He, and O2 has a total pressure of 1,015 kPa. Argon exerts a partial pressure of 152 kPa and Helium exerts a pa
arsen [322]
This one is easy. It's just addition and subtraction. Add the partial pressure of Argon and the partial pressure of Helium:
   152 kPa
<u>+305 kPa</u><u>
</u>  457 kPa

Next, subtract the result from the total pressure:

  1015 kPa
<u>-  457 kPa</u>
   558 kPa
8 0
3 years ago
A gas has a volume of 7.0 L and a mass of 8.50 X 109 grams. What is its density in g/ml?
natka813 [3]

Answer:

The answer is

<h2>1, 214, 285.71 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass of gas = 8.5 × 10^9 g

But 1 L = 1000 mL

So 7 L = 7000 mL

volume = 7000 mL

So the density of the gas is

density =  \frac{8.5 \times  {10}^{9} }{7000}  \\  = 1214285.714285...

We have the final answer as

1, 214, 285.71 g/mL

Hope this helps you

8 0
4 years ago
2C 2 H 2 (g)+5O 2 (g) CO 2 (g) + 2H 2 O (l)
Ganezh [65]

Answer:

2c2gh2+5cg2o4+2h2lo

Explanation:

7 0
3 years ago
Ethanol has a heat of vaporization of 38.56kj/mol and a normal boiling point of 78.4 ∘c.
Elanso [62]

This is an incomplete question, here is a complete question.

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 14 °C?

Answer : The vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

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 14.0^oC = ?

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

T_1 = temperature of ethanol = 14.0^oC=273+14.0=287K

T_2 = normal boiling point of ethanol = 78.4^oC=273+78.4=351.4K

\Delta H_{vap} = heat of vaporization = 38.56 kJ/mole = 38560 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})

P_1=5.174\times 10^{-2}atm

Hence, the vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

4 0
3 years ago
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