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Oksi-84 [34.3K]
3 years ago
7

you are holding a bowling ball with a mass of 6 kg at a height of 1 meter. How much gravitational potential energy does the bowl

ing ball have?
Chemistry
1 answer:
Bumek [7]3 years ago
7 0

The formula for GPE is PE=mgh, where “m” is the mass of the object, “g” is the acceleration due to gravity (~9.8 m/s^2 on Earth’s surface), and “h” is the height of the object from the ground. Therefore,

PE=mgh

PE=(6 kg)(9.8 m/s^2)(1 m)

PE=58.8 kg•m^2/s^2 or 58.8 Newtons

The GPE of the bowling ball under these conditions would be about 59 Newtons.

Hope this helps!

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N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?
just olya [345]

Answer:

\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     28.01               17.03

            N₂ + 3H₂ ⟶ 2NH₃

m/g:                          240.0

(a) Moles of NH₃

\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

(b) Moles of N₂

\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

(c) Mass of N₂

\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

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answer is A

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One of the assumptions states that "a gas is composed of a large number of identical molecules moving at different speeds".

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