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WARRIOR [948]
3 years ago
7

Suppose you mix 75 g of water at 15 ℃ with 25 g of water at 75 ℃. Predict the final temperature from the choices below. Explain

your choice.
A. 30 ℃ b. 45 ℃ c. 60 ℃
Chemistry
1 answer:
k0ka [10]3 years ago
7 0
It’s probs a because 75-25 is 30 so it’s probs 30°C
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For a solution of H2SO4, determine the van\'t Hoff factor assuming 0% ionization and 100% ionization
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When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is
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Answer:

38.83 %  of  CaCO3

61.17 %  of  MgCO3

Explanation:

where Moles of CaCO3 is equals to x and MgCO3 is y we have that...

CaCO3 molar mass = 100.09 g / mol  = 100.09 x

MgCO3 molar mass = 84.31 g / mol  = 84.31 y

decomposition reactions :

CaCO3 ---> CaO + CO2

MgCO3 ---> MgO + CO2  

So we have that , Moles of CaO = Moles of CaCO3 = x

and Moles of MgO = Moles of MgCO3 = y

CaO molar mass = 56.08 g / mol

MgO molar mass = 40.30 g / mol

CaO = 56.08 x

 MgO = 40.30 y

"If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3  

thus

56.08 x + 40.30 y = 0.51 ( 100.09 x + 84.31 y )

56.08 x + 40.30 y = 51.04 x + 42.99y

5.04 x = 2.7 y

y = 1.87 x    

CaCO3 % in the sample

= 100.09 x × 100 / ( 100.09 x + 84.31 y )

= 10009 x / ( 100.09 x + 84.31 × 1.87 x )

= 10009 x / ( x ( 100.09 + 157.66 ) )

= 10009 / 257.75

= 38.83 %

MgCO3 % in the sample

= 100 - 38.83  

=   61.17 %  

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A titration was performed on 50.0mL of 0.250 M NaOH.After the
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Answer:

1.750 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 7.14 mL

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Using the above formula , the molarity of acid , can be calculated as ,

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