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3 years ago

Volume of 8.29ml and a mass of 16.31g

Chemistry

1 answer:

vladimir2022 [97]3 years ago

3 0

If you wanna know density then:

Density = mass/volume = 16.31/8.29 = 1.96 g/mL.

Hope this helps!

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A 30-n net force on a skater produces an acceleration of 0.6 m/s2. what is the mass of the skater

kramer

Answer:

<h3>The answer is 50 kg</h3>

Explanation:

The mass of the skater can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{30}{0.6} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

Bromine pentachloride formula

nekit [7.7K]

Answer:

Bromine pentafluoride, BrF5

bromine pentafluoride is an interhalogen compound and a fluoride of bromine. It is a strong fluorination reagent. BrF5 finds use in oxygen isotope analysis.

4 0

3 years ago

The actual density of iron is 7.874 g/mL. In a laboratory investigation, Jason finds the density of a piece of iron to be 7.921

mixas84 [53]

The percent error associated with Jason’s measurement is 0.596%.

HOW TO CALCULATE PERCENTAGE ERROR:

The percentage error of a measurement can be calculated by following the following process:

Find the difference between the true value and the measured value of a quantity.
Then, divide by the true value and then multiplied by 100

The true value of the density of iron is 7.874 g/mL
Jason observed value is 7.921 g/mL

Difference = 7.921 g/mL - 7.874 g/mL

Difference = 0.047 g/mL

Percentage error = 0.047/7.874 × 100

Percentage error = 0.596%.

Therefore, the percent error associated with Jason’s measurement is 0.596%.

Learn more: brainly.com/question/18074661?referrer=searchResults

5 0

2 years ago

The rate at which a certain Australian tree cricket chirps is 194/min at 28°C, but only 47.6/min at 5°C, From these data calcula

uysha [10]

Answer: The energy of activation for the chirping process is 283.911 kJ/mol

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_2$ = rate of reaction at $28^0C$ = 194/min

$K_1$ = rate of reaction at $5^0C$ = 47.6 /min

$Ea$ = activation energy

R = gas constant = 8.314 J/Kmol

tex]T_1[/tex] = initial temperature = $5^oC=273+5=278K$

tex]T_1[/tex] = final temperature = $28^oC=273+28=301K$

Now put all the given values in this formula, we get

$\frac{194}{47.6}=\frac{E_a}{2.303\times 8.314}[\frac{1}{278}-\frac{1}{301}]$

${E_a}=283911J/mol=283.911kJ/mol$

Thus the energy of activation for the chirping process is 283.911 kJ/mol

8 0

3 years ago

How much oxygen will contain the same number of atoms as the number of molecules in 73 g of HCI?

expeople1 [14]

Answer:

64g of O2.

Explanation:

We'll begin by calculating the number of molecules in 73g if HCl.

This is illustrated below:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ molecules. This implies that 1 mole of HCl also contains 6.02×10²³ molecules

1 mole of HCl = 1 + 35.5 = 36.5g

Thus, if 36.5g of HCl contains 6.02×10²³ molecules, then 73g of HCl will contain = (73 x 6.02×10²³)/36.5 = 1.204×10²⁴ molecules.

Therefore, 73g of HCl will contains 1.204×10²⁴ molecules.

Now, we shall determine the mass of oxygen that will contain 1.204×10²⁴ molecules.

This can be obtained as follow:

1 mole of O2 = 16x2 = 32g

32g of O2 contains 6.02×10²³ molecules.

Therefore, Xg of O2 will contain 1.204×10²⁴ molecules i.e

Xg of O2 = (32 x 1.204×10²⁴)/6.02×10²³

Xg of O2 = 64g

Therefore, 64g of O2 will contain the same number of molecules (i.e 1.204×10²⁴ molecules) in 73g of HCl.

5 0

3 years ago