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xenn [34]
2 years ago
8

A relationship in which one variable increases as the other variable decreases is an inverse relationship.

Chemistry
1 answer:
Brut [27]2 years ago
8 0

Answer:

TRUE

Explanation:

HOPE IT WILL HELP YOU

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Devise a way to separate a mixture of sand, salt, and iron filings. Write the procedure using a step-by-step method and explain
aev [14]
1. Iron fillings are magnetic, so use a magnet to pull the iron fillings out of the mix.
2. Then you can put the salt and sand mixture into water, since salt is soluble, and the salt will dissolve, leaving you with sand.
3 0
3 years ago
the chloride of a metal M contains 47.25% of metal 1.0 gram of metal would be displaced from a compound by 0.88 gram of another
GenaCL600 [577]

Answer:

The equivalent weight of M is approximately 31.8 g

The equivalent weight of N is approximately 27.98 g

Explanation:

The given parameters are;

The percentage of the the metal M in in the chloride = 47.25%

Where by the chemical formula for the metal chloride is MClₓ, we have;

47.25% of the mass of MClₓ = Mass of M = W

Therefore, we have;

\dfrac{0.4725}{W} = \dfrac{1}{W + 35.5 \cdot x}

0.4725 × (W +  35.5·x) = W

0.4725·W + 0.4725×35.5×x = W

W - 0.4725·W  = 16.77·x

0.5275·W = 16.77·x

W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M

The equivalent weight of M = 31.799 ≈ 31.8 g

Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g

The equivalent weight of N = 27.98 g.

7 0
2 years ago
In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc
natita [175]

Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water 

x= 0.7535 g water




3 0
3 years ago
A solution is prepared by dissolving 27.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the
bagirrra123 [75]
The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is 
     % by mass = (mass of the solute / mass of the solution)*100 
                        = mass of solute / (mass of the solute + mass of the solvent)*100
                        = (27.7 g CaCl2 / 27.7g + 375g) * 100 
                        = 6.88
5 0
3 years ago
Read 2 more answers
A loaf bread has a mass of 500 g and volume of 12.0 cm3. what is the density of the bread
pychu [463]
41.6g/cm3 would be the density of the bread
5 0
3 years ago
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