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3 years ago

13

A student was given a piece of metal with a mass of 85.0 g. She placed it

1 answer:

Anastasy [175]3 years ago

6 0

Answer:

22mL

Explanation:

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What would happen if I swallowed a mento then drank some coca cola?

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Your stomach will bubble lol it might kind of tickle

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In which situation is no work considered to be done by a force?

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If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.

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Given the formula: what is a chemical name of this compound? propane propanal propanol propanone

katrin [286]

1) Chemical formula for propane is CH₃-CH₂-CH₃.
Propane is a three carbon alkane (acyclic saturated <span>hydrocarbon).
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Propanal <span> is a </span>saturated<span> three carbon </span>aldehyde (have<span> a </span>carbonyl<span> center).
3) </span>Chemical formula for propanol is CH₃-CH₂-CH₂-OH.
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4) </span>Chemical formula for propanone is (CH₃)₂-C=O.
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3 years ago

2. Calculate the mass of 5.35 mole of H2O2.

Dmitry_Shevchenko [17]

Answer:

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Explanation:

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2 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

2 years ago