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alukav5142 [94]
3 years ago
12

Knowing the gravity on the moon is 1.6m/s what is your weight?

Chemistry
1 answer:
wariber [46]3 years ago
3 0
Weight varies dramatically if we leave earths surface.On the moon for example acceleration due to gravity is only 1.67m/s2 A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the moon
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1. The
rodikova [14]

Answer:

1. Proton = electron if the element is not in an ionic state

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3 years ago
What is the pH of a solution that has a [H+] = 0.010 mol/L?
Serjik [45]

Answer:

pH = 2

A 0.010 M solution of hydrochloric acid, HCl, has a molarity of 0.010 M. This means that [H+] = 1 x 10-2 M. The pH of this aqueous solution of H+ ions is pH = 2

hope this helps :3

Download pdf
5 0
2 years ago
When a solution of AgNO3 is mixed with a solution of NaBr (multiple options could be correct):
alex41 [277]

Answer:

After a few minutes pass, the concentration of Ag+ and Br- will be lower than when the two solutions were first mixed.

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7 0
3 years ago
At 450°C, ammonia gas will decompose according to the following equation: 2 NH3 (g)  N2 (g) + 3 H2 (g) Kc = 4.50 at 475˚C An u
velikii [3]

Answer:

0.2024 M

Explanation:

For the decomposition reactio given, let's do an equilibrium chart. Let's call the initial concentration of NH₃ as C:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

C 0 0 Initial

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C - 2x x 3x Equilibrium

3x = 0.252

x = 0.084 M

The equilibrium constant (Kc) is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of reactants concentrations elevated by their coefficients.

Kc = ([H₂]³*[N₂])/([NH₃]²)

4.50 = [(0.252)³*(0.084)]/(C - 2*0.084)²

4.50 = 0.00533/(C - 0.168)²

4.50 = 0.00533/(C² - 0.336C + 0.028224)

4.50C² - 1.512C + 0.127008 = 0.00533

4.50C² - 1.512C + 0.121678 = 0

Solving the equation by a graphic calculator, for C > 0.168

C = 0.2024 M

4 0
3 years ago
What does stoichiometry use to relate moles of one molecule to moles of
Shkiper50 [21]
It is A I took it befor hope this helps:)
6 0
2 years ago
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