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3 years ago

how many moles of solute are contained in 500.00 milliliter solution that has a concentration of 0.35M

Chemistry

1 answer:

igomit [66]3 years ago

6 0

1.75 moles of Solute or 1.75n.

To find this, you'll need to use the molarity equation which is M = $\frac{n}{v}$

List the given items to organize better:

M (Molarity) = .35

V (Volume) = 5L [must always be in Liters]

n (Moles) = ?

Now, for the equation, substitute the variables with the given numbers:

.35 = $\frac{n}{5}$

Isolate the variable by multiplying the volume, 5, onto both sides. This will give you the answer of:

1.75 = n

Hope this helps!

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3 years ago

Car manufacturers are developing engines that use H₂ as fuel. In Iceland, Sweden, and other parts of Scandinavia, where hydroele

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If the coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.

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(b) Given

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It is expressed as

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= $\frac{(12)(3.5.10^{6} )}{(0.082057)(298)}$

= 1.7176 ×10⁶ mole

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= 2 × 1.7176 × 10⁶ moles of electron

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We know that

charge = moles of electron × F

= (3.435 ×10⁶ × 96500) C

= 3.31 × 10¹¹ C

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1.44 V = $\frac{energy}{3.31 . 10^{11} }$

Energy = 1.44 V × 3.31 × 10¹¹ C

= 4.766 × 10¹¹ Joules

Thus from the above conclusion we can say that, coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.

Learn more about Fuel Cell here : brainly.com/question/16612142

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Question : Car manufacturers are developing engines that use H₂ as fuel. In Iceland, Sweden, and other parts of Scandinavia countries, where hydroelectric plants produce inexpensive electric power, the H₂ can be made industrially by the electrolysis of water.

a) How many moles of electrons flow per mole of reaction ?

b) How many coulombs are needed to produce 3.53X10⁶ L of H₂ gas at 12.0 atm and 25°C.

c) If the coulombs are supplied at 1.44 V, how many joules are produced?

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2 years ago

The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat

guajiro [1.7K]

First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

= 1 mol * 18 mol/g

= 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

= 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

= 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

= 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

= 1.881 KJ +6.01 KJ + 4.514 KJ

= 12.405 KJ

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3 years ago