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bogdanovich [222]
3 years ago
5

Express the following numbers as standard form: a) 1.52 x 10^-2 b) 7.78 x 10^-8

Chemistry
1 answer:
Vikentia [17]3 years ago
4 0

Answer:

0.0152

0.0000000778

Explanation:

You just need to more the decimal as many times as the exponent to the to the left if it is negativa and to the right if it is positive.

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The diagram below shows the different phase transitions that occur in matter. Three bars are shown labeled Solid, Liquid, and Ga
Grace [21]

The arrow that will represent the phase change that involves the same amount of energy as arrow 1 will be arrow 4.

<h3>Phase change</h3>

Arrow 1 represents a phase change from liquid to gas while arrow 4 represents a phase change from gas to liquid.

In other words, arrow 1 and arrow 4 are direct opposites of one another,

This means that if X amount of energy is required for arrow 1, the same amount of energy will be needed for arrow 4 but in the reverse direction.

More on phase change can be found here: brainly.com/question/12390797

#SPJ1

5 0
2 years ago
Help! giving brainly if correct​
Gnesinka [82]

Answer:

I would say the answer is b.

Explanation:

6 0
2 years ago
Read 2 more answers
B) How many kilojoules of heat will be released by the combustion of 22.52 g of this liquid at
snow_tiger [21]

Answer:

Explanation:

You realize that C2H5OH releases -1277.3kJ/mol. We need to convert this to the amount based on the question. We that 22.52g of C2H5OH = 0.48884 mol.

This means that it will release (-1277.3)(0.48884) = 624.40 KJ of heat will be released. Note the negative sign is not necessary here (I think) because it says how much is released and not the change in heat of the system so it should be positive.

3 0
3 years ago
Benzoic acid, c6h5cooh, has a ka = 6.4 x 10-5. what is the concentration of h3o+ in a 0.5 m solution of benzoic acid? 3.2 x 10-5
FinnZ [79.3K]
The answer for this issue is: 
The chemical equation is: HBz + H2O <- - > H3O+ + Bz- 
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz] 
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x. 
Accept that x is little contrasted with 0.5 M. At that point, 
Ka = 6.4X10^-5 = x^2/0.5 
x = [H3O+] = 5.6X10^-3 M 
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make) 
3 0
4 years ago
A 25.00 mL sample of the ammonia solution
musickatia [10]

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 =  250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

3 0
3 years ago
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