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2 answers:

8 0

6. False, Gravity is a force that cannot be changed
7. False, an object with more mass has more/less gravitational force than an object with a smaller mass. Objects that are closer together have more/less of a gravitational force between them than objects that are further apart. The more separation two objects have, the more the force between them decreases.
8. True, objects that are closer together have more/less of a gravitational force between them than objects that are further apart. The more separation two objects have, the more the force between them decreases. When mass 1 increases the force between the objects increases.
9. False, Jupiter, the fifth planet from the Sun, has the strongest gravitational pull because it's the biggest and most massive.

I hope this helps!!

qaws [65]3 years ago

4 0

The answer is 6f 7t 8f 9f

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B. The equilibrium constant for the reaction 2H2(g) + S2(g) 2H2S(g) is Keq = . (3 points)

eimsori [14]

Answer:

i. Keq=4157.99.

ii. More hydrogen sulfide will be produced.

Explanation:

Hello,

i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:

$Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99$

ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.

Best regards.

4 0

3 years ago

A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance

sasho [114]

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$ , where K is the Coulomb´s constant ( $9x10^{9} N\frac{m^{2} }{C^{2} }$ ), $q_{1}$ = Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$ .

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$