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oksian1 [2.3K]
3 years ago
9

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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Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re
bearhunter [10]

Answer:  28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}   

\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles

The balanced chemical equuation is:

4Al+3O_2\rightarrow 2Al_2O_3  

According to stoichiometry :

4 moles of Al produce == 2 moles of Al_2O_3

Thus 0.556 moles of Al will produce=\frac{2}{4}\times 0.556=0.278moles  of Al_2O_3

Mass of Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0
3 years ago
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vovangra [49]

Answer:

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Explanation:

6 0
3 years ago
Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°
Pavel [41]

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that  involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

PV = \frac{m}{M}RT

PxM = \frac{m}{V}RT

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

2.7 = \frac{0.94xMavg}{0.082x289}

0.94Mavg = 63.9846

Mavg = 68.0687 g/mol

The molar mass of N is 14 g/mol and of O is 16 g/mol, than M_{NO2} = 46 g/mol and M_{N2O4} = 96 g/mol. Calling y the molar fraction:

Mavg = M_{NO2}y_{NO2} + M_{N2O4}y_{N2O4}

And,

y_{NO2} + y_{N2O4} = 1

y_{N2O4} = 1 - y_{NO2}

So,

68.0687 = 46y_{NO2} + 92x(1 - y_{NO2})

68.0687 - 92 = 46y_{NO2} - 92y_{NO2}

46y_{NO2} = 23.9313

y_{NO2} = 0.52

y_{N2O4} = 0.48

The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm

8 0
3 years ago
Identify the products in the following chemical reaction: Multiply chose
rewona [7]

Answer:

C. CaCO3

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3 years ago
How many significant figures are in the number 420,000 mi?
Mnenie [13.5K]

Answer:

2

Explanation:

zeros are not considered sig figs

3 0
3 years ago
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