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Oxana [17]
2 years ago
5

A mixture of potassium bromide and potassium hydroxide has a total mass of 5.50 g. If the mixture contains 2.45 g K, what is the

mass in grams of potassium hydroxide in the mixture?
Chemistry
1 answer:
Zigmanuir [339]2 years ago
5 0
KBr molar mass = 39 + 80 = 119 grams which has x moles in the mixture
KOH molar mass = 39 + 17 = 56 grams which has y moles in the mixture

119x + 56 y = 5.50
39*x + 39*y = 2.45 that's because 

Every mole of KOH and KBr has exactly 1 mole of K in each molecule.

I won't go through the math, but the answer is
x = 0.0315
y = 0.0314
The answer to your question involves y*56 from which we get 1.758 grams Sodium Hydroxide

To check we
119x + 56y = 5.50 grams of mixture.
119* 0.0315 = 3.75 grams
56 * 0.0324 = 1.758
Total = 5.51 which is about as close as you can get when rounding.
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Answer:

The answer to your question is: 1, 2, 1, 2

Explanation:

                       1 Fe(s)  + 2 Na⁺(aq)  → 1 Fe²⁺(aq)  + 2 Na(s)

                             Fe⁰   -   2e⁻       ⇒           Fe⁺²        Oxidases

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2 years ago
The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t
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The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.
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What mitigation measures can communities do to reduce the damage and impact of sudden geologic hazards?
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Explanation:

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2 years ago
What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its
pishuonlain [190]

Answer:

c = 4016.64 j/g.°C

Explanation:

Given data:

Mass of substance = 2.50 g

Calories release = 12 cal (12 ×4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 20°C - 25°C

ΔT  = -5°C

50208 j = 2.50 g . c. -5°C

50208 j = -12.5 g.°C .c

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