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2 years ago

13

Two balloons are charged with an identical quantity and type of charge: 0.004 C. They are held apart at a separation distance of

5m. Determine the magnitude of the electrical force of repulsion between them.

1 answer:

lukranit [14]2 years ago

6 0

Answer:

The answer to your question is F = 28800 N

Explanation:

Data

q₁ = q₂ = 0.004 C

distance = r = 5 m

k = 9x 10⁹ C

Force = F

Formula

F = k q₁ q₂ / r²

-Substitution

F = (9 x 10⁹)(0.004)(0.004) / (5)²

-Simplification

F = 144000 / 25

-Result

F = 28800 N

-Conclusion

The force of repulsion between these balloons is 28800 N

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AlekseyPX

Calcium ions have oxidation state 2+ => Ca (2+).

Bromime ions (bromide) have oxidation state 1- => Br (-).

So, to be neutral the compound has to have two Br (-) ions per each Ca(2+) ion.

That is represented in the chemical formula as Ca Br2, where the number 2 to the right of Br is a subscript meaning that there are two atoms of Br per each atom of Ca (the lack of subscript means 1 atom).

Answer: Ca Br2.

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2 years ago

Read 2 more answers

If the concentration of a saturated solution at 0∘C is 12.5 gCuSO4/100 g soln, what mass of CuSO4⋅5H2O would be obtained? [Hint:

Andrew [12]

Answer:

Mass of CuSO4.H2O obtained: $m_{total}=19.52 g$

Explanation:

The molecular weight of the salt is: $M=159.5 g/mol$

<u>In the solution</u>: 12.5 g of CuSO4

In moles: $n=\frac{12.5 g}{159.5 g/mol}=0.078 mol$

<u>Mass of wate</u>r:

5 moles of water per mol of salt: $m=\frac{5mol}{1mol}*0.078mol*\frac{18g}{mol}=7.02 g$

Mass of CuSO4.H2O obtained: $m_{total}=12.5 g + 7.02 g=19.52 g$

7 0

2 years ago

If 200mL of 0.60 M MgCl2 (aq) is added to 400mL of distilled water, what is the concentration of Mg2+(aq) in the resulting solut

lara [203]

Answer:

A.0.20M

Explanation:

c 1 V 1 = c 2 V 2

Initial Volume, V1 = 200 mL

Final Volume, V2 = 200 + 400 = 600 mL

Initial Concentration, c1 = 0.60 M

Final Concentration, c2= ?

Solving for c2;

c2 = c1v1 / v2

c2 = 0.60 * 200 / 600

c2 = 0.20M

3 0

1 year ago

A sample of SO3 is introduced into an evacuated sealed container and heated to 600 K. The following equilibrium is established:

Andreas93 [3]

Answer: The value of $K_p$ is 0.050.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_x=x\times P$

As we know the mole fraction of $O_2$ is 0.12

The partial pressure of $O_2=0.12\times 3.0atm=0.36atm$

The partial pressure of $SO_2=2\times 0.36atm=0.72atm$ Thus the partial pressure of $SO_3$ is = [3 - (0.36+0.720)] atm = 1.92 atm

$p_{SO3}$ = 1.92 atm

$2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$

$K_p=\frac{p_{O_2}\times (p_{SO_}2)^2}{(p_{SO_3})^2}$

$K_p=\frac{0.36\times (0.72)^2}{(1.92)^2}$

$K_p=0.050$

The value of $K_p$ is 0.050.

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2 years ago

9 The Haber process is a reversible reaction. N2(g) + 3H2(g) 2NH3(g) The reaction has a 30% yield of ammonia. Which volume of am

Illusion [34]

Answer: 3.36 L of ammonia gas

Explanation:

The balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

According to stoichiometry :

3 moles of $H_2$ produce = 2 moles of $NH_3$

Thus 0.75 moles of $H_2$ will producee= $\frac{2}{3}\times 0.75=0.50moles$ of $NH_3$

But as percent yield is 30 %, amount of ammonia produced = $\frac{30}{100}\times 0.50moles=0.15moles$

According to ideal gas equation:

$PV=nRT$

P = pressure = 1 atm

V = Volume = ?

n = number of moles = 0.15

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$

$V=\frac{nRT}{P}$

$V=\frac{0.15\times 0.0820 L atm/K mol\times 273K}{1atm}=3.36L$

Thus 3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.

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1 year ago