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3 years ago

Density of iron is 7.8 g/cc. How much would 10 cc of iron weigh?

Chemistry

1 answer:

Elena-2011 [213]3 years ago

4 0

Answer:

<h3>The answer is 78 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 10 cc

density = 7.8 g/cc

So we have

mass = 7.8 × 10

We have the final answer as

Hope this helps you

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3 years ago

Assume an object A with a mass of 66.789 g and a volume of 10.1 mL. what is its density?

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Answer: 6.61

Explanation:

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3 years ago

How many particles are present in0.24moles of carbon?

Ray Of Light [21]

Answer:

1.45 x 10²³ particles

Explanation:

Given parameters:

Number of moles of carbon = 0.24moles

Unknown:

Number of particles = ?

Solution:

A mole of a substance contains the Avogadro's number of particles.

The Avogadro's number of particles is 6.02 x 10²³

So;

0.24 moles of carbon will contain 0.24 x 6.02 x 10²³ = 1.45 x 10²³ particles

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2 years ago

Iron will most likely react with which of the following elements?

gtnhenbr [62]

Answer:

bromine (Br)

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3 years ago

A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new

Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

$PV=nRT$

P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

$V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L$

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

$PV=nRT$

P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

$n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles$

Thus the now the container contains 2.07 moles.

6 0

2 years ago