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3 years ago

A solution contains 32.7 g H3PO4 in 455 mL of solution. Calculate its molarity

Chemistry

1 answer:

Aleks [24]3 years ago

4 0

Morality= wt\m.wt*1000\v(ml)

So the m.wt of h3po4 is 98.00g\ml
Molarity = 32.7g\98.00*1000\455ml
=0.7176

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The measurement 0.0002833g has how many significant figures?

kherson [118]

Answer:

There are seven significant figures

Explanation:

There are seven different digits within the number. Three 0s, one 2, one 8, and two 3s, adding up to seven different numbers. You exclude the first 0 when the number is a decimal, leaving seven significant figures. Hope this makes sense! :)

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-FeS2+-O2->-Fe2O3+-SO2 What coefficient should be placed in front of SO2 to balance the equation?

NISA [10]

Answer:

8 should be placed in front of SO₂.

Explanation:

The easiest way to solve this question is by writing the <u>entire</u> balanced equation:

4FeS₂ + 11O₂ -> 2Fe₂O₃ + 8SO₂

We can achieve this by first balancing the Fe, then S, and finally the O.

We can also double check our answer by counting the number of each element on both sides:

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3 years ago

A 32.78 g sample of a substance is initially at 22.7 °c. after absorbing 2017 j of heat, the temperature of the substance is 173

larisa86 [58]

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3 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

<u>Where:</u>

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

4 0

3 years ago