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Iteru [2.4K]
3 years ago
6

Problems: Upload your work for all 5 questions when you are on the last question.

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

1.08 M

Explanation:

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State the law of conservation of mass
Katen [24]

It’s basically that’s any system that’s closed to all transfers of matter and energy the mass of the system has to remain constant over time because they can’t change meaning you can’t add or remove from it

8 0
3 years ago
Suppose you were to quickly throw a ball in outer space. Would it slow down as quickly in space, as it does in Earth? Explain wh
Sholpan [36]

Answer:It would never stop until something hit the ball, to slow it down.

Explanation:

This is so because there is no gravitational pull in space.

7 0
3 years ago
Please answer asap MUCH NEEDED!!
gulaghasi [49]

Answer:

I think its C I am sorry if I am wrong

3 0
3 years ago
An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.
aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

6 0
3 years ago
What is my theoretical yield (in moles) of Potassium Bromide (KBr) if I start with 40 grams of Iron (II) Bromide [FeBr2]? moles
Verizon [17]
The reaction will be: FeBr2 + K --> KBr + Fe
Balancing gives: FeBr2 + 2K --> 2KBr + Fe
The molar mass of FeBr2 is 55.85 + 2*79.9 = 215.65 g/mol.
We divide 40 g / 215.65 g/mol = 0.185 mol FeBr2
Based on stoichiometry:
(0.185 mol FeBr2)(2 mol KBr/1 mol FeBr2) = 0.370 mol KBr
4 0
4 years ago
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