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Sladkaya [172]
3 years ago
10

Joe is the best baseball player in the world. He can throw a baseball faster than anyone. If the mass of the baseball was increa

sed by 50%, but he applied the same force as he did to a lighter baseball, which of these would best describe Joe's attitude? A)Joe would love to use the new ball in future baseball games.
B) Joe would be happy as the new ball would have a faster acceleration.
C) Joe would be unhappy as the new ball would have a slower acceleration.
D) Joe's attitude would not change as there would be no change in the acceleration.
Chemistry
2 answers:
GarryVolchara [31]3 years ago
8 0

We know tha

Force = Mass X acceleration

acceleration = this provides the distance to the base ball

If mass of baseball is increased keeping force constant, the acceleration will decrease, this will decrease the distance traveled by the base ball

so Joe would be unhappy as the new ball would have a slower acceleration.

storchak [24]3 years ago
4 0
Try C that is what im going with
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A large sport utility vehicle has a mass of 2600 kg.
Pepsi [2]
<span>Mass of CO2 = 225.632g</span>
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Which one of the following statements is true?
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Answer:

A calculator has an endifiite shape because all of its atoms are touching each other

8 0
3 years ago
If you burn 55.6 g of hydrogen and produce 497 g of water, how much oxygen reacted?
tester [92]

Answer:

441.28 g Oxygen

Explanation:

  • The combustion of hydrogen gives water as the product.
  • The equation for the reaction is;

2H₂(g) + O₂(g) → 2H₂O(l)

Mass of hydrogen = 55.6 g

Number of moles of hydrogen

Moles = Mass/Molar mass

          = 55.6 g ÷ 2.016 g/mol

          = 27.8 moles

The mole ratio of Hydrogen to Oxygen is 2:1

Therefore;

Number of moles of oxygen = 27.5794 moles ÷ 2

                                               = 13.790 moles

Mass of oxygen gas will therefore be;

Mass = Number of moles × Molar mass

Molar mass of oxygen gas is 32 g/mol

Mass = 13.790 moles × 32 g/mol

<h3>          = 441.28 g</h3><h3>Alternatively:</h3>

Mass of hydrogen + mass of oxygen = Mass of water

Therefore;

Mass of oxygen = Mass of water - mass of hydrogen

                          = 497 g - 55.6 g

<h3>                           = 441.4 g </h3>
6 0
4 years ago
Read 2 more answers
Hydrazine (N 2 H 4 )) a rocket fuel reacts with oxygen to form nitrogen gas and water vapor . The reaction is represented with t
Drupady [299]

The mass of hydrazine (N₂H₄) required to produce 96 g of water (H₂O) is 85.4 g (Option C)

<h3>Balanced equation </h3>

N₂H₄ + O₂ —> N₂ + 2H₂O

Molar mass of N₂H₄ = (2×14) + (4×1) = 32 g/mol

Mass of N₂H₄ from the balanced equation = 1 × 32 = 32 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

<h3>How to determine the mass of N₂H₄</h3>

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

Therefore,

96 g of H₂O will be produced by = (96 × 32) / 36 = 85.4 g of N₂H₄

Thus, 85.4 g of N₂H₄ is needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

8 0
2 years ago
1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
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