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Alja [10]
3 years ago
6

How close does Venus get to earth

Chemistry
1 answer:
Sergio [31]3 years ago
4 0
<span>261 million kilometers</span>
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Which of the following is a chemical change
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C. a burning candle

all the other choices are physical changes

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Which is an IUPAC name for a covalent compound?. . A. carbon dioxide. B. carbon bioxide. C. zinc monophosphate. D. zinc phosphat
Mumz [18]
<span>Which is an IUPAC name for a covalent compound?. . A. carbon dioxide</span>
8 0
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Explain why the atmosphere is classified as a mixture
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The components are not permanently together. They can be separated and thats why its a mixture.
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You have 34.1 x 1023 molecules of O2. How many moles of O2 do you have? Fill in the grid. Then, solve the problem and answer the
denis-greek [22]

Number of moles = 5.7 moles of oxygen.

<u>Explanation:</u>

We have to convert number of molecules into number of moles by dividing the number of molecules by Avogadro's number.

Here number of molecules of oxygen given is 34.1 × 10²³ molecules.

Now we have to divide the number of molecules by Avogadro's number as,

Number of moles = $\frac{number of molecules}{Avogadro's number}

                            = $\frac{34.1}{6.022} \times \frac{10^{23} }{10^{23} }

                          = 5.7 moles

So here molecules is converted into moles.

5 0
3 years ago
A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion
tensa zangetsu [6.8K]

Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

Heat of combustion = heat capacity of calorimeter * deltaT\\

Given,

The heat of combustion = 10.5kJ/g.

deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

deltaT =heat  capacity of calorimeter   * (change in temperature)\\10.5kJ/g = heat  capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC

Answer:

The heat capacity of the calorimeter is 3.0kJ/g.^oC.

5 0
3 years ago
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