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cestrela7 [59]
3 years ago
11

A force of 350 N causes a body to move with an acceleration of 10 m/s2. What's the mass of the body?

Chemistry
2 answers:
nirvana33 [79]3 years ago
5 0

your answer is 35kg

i just took the test

Scrat [10]3 years ago
3 0
You have to use the equation F=ma and solve for m to get m=F/a.
m=mass in kg
F=force (in this case 350N)
a=acceleration (in this case 10m/s²)
when you plug everything in you should find that m=35kg

I hope this helps.
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A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure o
larisa86 [58]

Answer:

  • <em>The partial pressure of oxygen in the mixture is</em><u> 320.0 mm Hg</u>

Explanation:

<u>1) Take a base of 100 liters of mixture</u>:

  • N: 60% × 100 liter  = 60 liter

  • O: 40 % × 100 liter = 40 liter.

<u>2) Volume fraction:</u>

At constant pressure and temperature, the volume of a gas is proportional to the number of molecules.

Then, the mole ratio is equal to the volume ratio. Callin n₁ and n₂, the number of moles of nitrogen and oxygen, respectively, and V₁, V₂ the volume of the respective gases you can set the proportion:

  • V₁ / V₂ = n₁ / n₂

That means that the mole ratio is equal to the volume ratio, and the mole fraction is equal to the volume fraction.

Then, since the law of partial pressures of gases states that the partial pressure of each gas is equal to the mole fraction of the gas multiplied by the total pressure, you can draw the conclusion that the partial pressure of each gas is equal to the volume fraction of the gas in the mixture multiplied by the total pressure.

Then calculate the volume fractions:

  • Volume fraction of a gas = volume of the gas / volume of the mixture

  • N: 60 liter / 100 liter = 0.6 liter

  • V: 40 liter / 100 liter = 0.4 liter

<u>3) Partial pressures:</u>

These are the final calculations and results:

  • Partial pressure = volume fraction × total pressure

  • Partial pressure of N = 0.6 × 800.0 mm Hg = 480.0 mm Hg

  • Partial pressure of O = 0.4 × 800.0 mm Hg = 320.0 mm Hg
8 0
3 years ago
Which statement explains why a C–O bond is
Nataly_w [17]

Answer: (3) The difference in electronegativity between  carbon and oxygen is greater than that  between fluorine and oxygen.

Explanation: Polarity of a molecule is due to the difference in electronegativity of the atoms. More is the electronegativity difference, more is the polarity.

Electronegativity of carbon = 2.5

Electronegativity of oxygen = 3.5

Electronegativity of fluorine = 4.0

Thus the difference in electronegativity of carbon and oxygen is=(3.5-2.5)= 1.0

Thus the difference in electronegativity of fluorine and oxygen is=(4.0-3.5)= 0.5.

Thus C-O bond is more polar than F-O bond.

4 0
3 years ago
Read 2 more answers
How many moles are in 564 grams of Copper
nexus9112 [7]
63.55 is how many miles are in 564 grams of copper
4 0
3 years ago
The total pressure of gases a, b, and c in a closed container is 4.1 atm. if the mixture is 36% a, 42% b, and 22% c by volume, w
Natali [406]

 The  partial  pressure of  gas C  is 0.902  atm


  calculation

partial pressure of gas c  =[( percent by volume of  gas   C /  total  percent)   x total pressure]


percent  by  volume of gas C= 22%

Total   percent  = 36% +42%  + 22%  = 100 %

Total  pressure  =  4.1 atm


partial  pressure  of gas C  is therefore =  22/100 x 4.1 atm = 0.902  atm

3 0
3 years ago
Read 2 more answers
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
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