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3 years ago

A bowling ball is pushed with a force of 22.0 N and accelerates at 5.5 m/s^2 what is the mass of the bowling ball?

Chemistry

1 answer:

Kruka [31]3 years ago

7 0

Data:

F (force) = 22.0 N

m (mass) = ? (in Kg)

a (acceleration) = 5.5 m/s²

We apply the data to the Resultant Force formula, we have:

4.0 kg

I hope this helps. =)

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Answer:

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2 years ago

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statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

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3 years ago

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I think the answer for this is 4702.5 J/g*k Depending on if it is water as a solid liquid or gas. I used water as a liquid when I solved it. J=(75g)(4.18 J/g*k)(15K)

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3 years ago

Select all that apply.

cupoosta [38]

Answer:

B and C

Explanation:

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2 years ago

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Answer:

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