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yarga [219]
2 years ago
15

How many grams of solute are needed to make 2.50 L of a 1.75 M solution of Ba(NO3)2?

Chemistry
1 answer:
Eddi Din [679]2 years ago
3 0

1143.4grams of solute are needed to make 2.50L of a 1.75M solution of Ba(NO3)2. Details about molarity can be found below.

<h3>How to calculate mass?</h3>

The mass of a substance can be calculated by multiplying the number of moles by its molar mass.

However, the number of moles can be calculated by using the following formula:

Molarity = no of moles ÷ volume

no of moles = 1.75 × 2.5

no of moles = 4.38mol

molar mass of Ba(NO3)2 = 261.34g/mol

mass of Ba(NO3)2 = 261.34 × 4.38

mass of Ba(NO3)2 = 1143.4grams.

Therefore, 1143.4grams of solute are needed to make 2.50L of a 1.75M solution of Ba(NO3)2.

Learn more about mass at: brainly.com/question/19694949

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5 0
3 years ago
consider the following equilibrium:h2co3(aq) h2o(l) h3o (aq) hco3-1(aq).what is the correct equilibrium expression?
MAVERICK [17]

The equilibrium expression shows the ratio between products and reactants. This expression is equal to the concentration of the products raised to its coefficient divided by the concentration of the reactants raised to its coefficient. The correct equilibrium expression for the given reaction is:<span>

<span>H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)

Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>

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3 years ago
What is the temperature -71 oC expressed in Kelvin?<br> Group of answer choices
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Explanation:

-71°C + 273.15

= 202.15K

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4 years ago
What does the amount of kinetic energy an object has depend on?
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EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK
mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = 5.0 X 10^{-4} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}

alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}

= 2.02 X 10^{-13}

When,

pH = -log[H+] = 10.15

[H+] = 7.08 X 10^{-11} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^-11

alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}

= 5.12 X 10^{-23}

4 0
3 years ago
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