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Elina [12.6K]
3 years ago
5

A reaction has ∆H = −356 kJ and ∆S = −36 J/K. Calculate ∆G (kJ) at 25°C.

Chemistry
1 answer:
Cloud [144]3 years ago
5 0

Answer: -345.2 KJ

Explanation: As we know that ,dG=dH-TdS

T=25+273=298 K

dG= -356 x1000-298(-36)= -356000+10728

=-345272 j

= -345.2 KJ

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Sergio039 [100]
The correct name of the compound Mn3(PO4)2 is  definitely the last option represented above <span>D. manganese(II) phosphate. I am pretty sure this answer will help you

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3 years ago
Arrange the following compounds from left to right in order of increasing percentage by mass of hydrogen: a) H2O, b) C12H26, c)
Nuetrik [128]

Answer:

(A) N4H6  (B) H2O  (C) LiH  (D) C12H26

Explanation:

The given compounds have been arranged  from left to right in order of increasing percentage by mass of hydrogen.

The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.

4 0
3 years ago
Water is dripping into a bucket at a rate of 51 drops per minute. if each drop has a volume of 0.050 milliliters, what volume of
serious [3.7K]

112.2 milliliters volume of water will drip into the bucket in 44 minutes.

<h3>What is volume?</h3>

How much space an object or substance takes up. • Measured in cubic meters (m3), liters (L) & milliliters (mL).

Total drop = Drops per minute X time

= 51 drops per minute X 44 minutes

=2244 drop

Volume of water will drip into the bucket in 44 minutes

=Total drop X Volume of each drop

=0.050 milliliters X 2244 drop

=112.2 milliliters

Hence, 112.2 milliliters volume of water will drip into the bucket in 44 minutes.

Learn more about volume here:

brainly.com/question/10904074

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Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.
Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

<em>Where [] are concentrations in equilibrium</em>

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized  over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>
8 0
3 years ago
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