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snow_lady [41]
3 years ago
14

If the density of a metal bar is 3.47 g/mL and volume is 30 mL, calculate the mass of the metal bar in grams (g). Show the steps

to answering this equation.
Chemistry
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

m=104.1g

Explanation:

Hello,

In this case, considering that the density is considered as the quotient of the mass and volume:

\rho =\frac{m}{V}

We can easily compute the mass by solving for it as follows:

m=V*\rho =30 mL*3.47g/mL\\\\m=104.1g

Best regards.

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How many grams of KCl 03 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?
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11.48-gram of KCl0_3 are needed to produce 6.75 Liters of O_2  gas measured at 1.3 atm pressure and 298 K

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

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P= 1.3 atm

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n=?

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T=298 K

Putting value in the given equation:

\frac{PV}{RT}=n

n= \frac{1.3 \;atm\; X \;6.75 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 298}

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0.3588 moles = \frac{mass}{32}

Mass= 11.48 gram

Hence, 11.48-gram of KCl0_3 are needed to produce 6.75 Liters of O_2 gas measured at 1.3 atm pressure and 298 K

Learn more about the ideal gas here:

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