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3 years ago

PLEASE HELP ASAP

Chemistry

1 answer:

sergij07 [2.7K]3 years ago

7 0

Both liquids and gasses are fluids is the answer

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Given the balanced equation 2KC103+ 2KC1+302

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is it decomp single replacement double replacement

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2 years ago

Which of these represents a compound ? * Water Hydrogen Oxygen

fiasKO [112]

Answer:

water

Explanation:

H2O

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3 years ago

Read 2 more answers

What is the percent composition of hydrogen if a sample is found to contain 7 grams of hydrogen, 32 grams of nitrogen, and 7 gra

nata0808 [166]

Answer: The percent composition of hydrogen in the sample is 15.22 %

Explanation:

We are given:

Mass of hydrogen = 7 grams

Mass of nitrogen = 32 grams

Mass of carbon = 7 grams

Total mass of the sample = 7 + 32 + 7 = 46 grams

To calculate the percentage composition of hydrogen in sample, we use the equation:

$\%\text{ composition of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of sample}}\times 100$

Mass of sample = 46 g

Mass of hydrogen = 7 g

Putting values in above equation, we get:

$\%\text{ composition of hydrogen}=\frac{7g}{46g}\times 100=15.22\%$

Hence, the percent composition of hydrogen in the sample is 15.22 %

6 0

3 years ago

The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr - Cp*mu/k where Cp is the heat

algol [13]

Answer:

The Prandtl number for this example is 14,553.

Explanation:

The Prandlt number is defined as:

$Pr=\frac{C_{p}*\mu}{k}$

To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.

$\mu=1896 \frac{lbm}{ft*h}*\frac{1000 g}{2.205 lbm}*\frac{3.281 ft}{1 m}*\frac{1h}{3600s} =7938 \frac{g}{m*s}$

Now that we have coherent units, we can calculate Pr

$Pr=\frac{C_{p}*\mu}{k}=0.66*7938/0.36=14553$

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3 years ago

What is the effect of an insoluble impurity, such as sand, on the observed melting point of a compound?]?

Taya2010 [7]

The effect of an insoluble impurity, such as sand, on the observed melting point of a compound would be none. It will not depress or elevate the melting point of the compound. Instead, it would affect the reading if you are trying to determine the melting point of the compound. This is because you might be missing the actual melting point of the compound since you will be waiting for the whole sample to liquify. You would not be able to determine exactly that temperature because of the insoluble impurity would have a different melting point than that of the compound.

7 0

3 years ago