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amm1812
4 years ago
7

Total mass of H2O2 in 20 grams of an aqueous H2O2 solution

Chemistry
1 answer:
sergeinik [125]4 years ago
7 0

Answer:

3=  (x/20.0g) x 100%

or    (20)(0.03)

Answer  0.6g

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A sample of 250 g of water are heated from 40°C to 121°C, calculate the amount of heat energy absorbed.
Marysya12 [62]

Answer:

the anwser isn't in the choices

Explanation:

H=MC(change of temp.)

M=mass of water=250g

C=specific heat of water = 4.186 j/g

change in temperature is 121-40= 81

H= 250x4.186x81=84766.5J

7 0
3 years ago
One mole of ice at 0°C is added to two moles of water at 50°C under a constant external pressure of 1 atm. This process is carri
kotykmax [81]

Answer:

T2 = 29.79°C

Explanation:

Equliibrium signifies that heat loss = heat gained

Heat gained by Ice;

H = ML

Mass, M = Number of moles * Molar mass = 1 * 18 = 18g

l = 6.01 k J m o l = 334 J/g

C = 4.186 J/g

H =  18(334)

H = 6012

Heat lost by water

H = MCΔT

H = 18 * 4.186 * (50 - T2)

H = 3767.4 - 75.348T2

Since H = H, we have;

6012 = 3767.4 - 75.348T2

- 75.348T2 = 3767 - 6012

T2 = 2245 / 75.348

T2 = 29.79°C

8 0
3 years ago
How many moles of each element (C, H, and O) are present in a 100.0 g sample of<br> ethanol?
Anuta_ua [19.1K]

Answer:

2.173 moles of ethanol is presented in a 100.0g sample of ethanol .

Explanation:

The amount of substance that contains as many Particles as there are atoms in exactly 12g of carbon- '12 isotope is called 1 mole '= 46 u.

7 0
3 years ago
has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a
Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

               A + B \rightarrow C

Initial :     0.3   0.4          0

Change:  -x       -x           x

Equilbm: (0.3 - x)  (0.4 - x)  x  

We know that, relation between standard free energy and equilibrium constant is as follows.

      \Delta G = -RT ln K

Putting the given values into the above formula as follows.

      \Delta G = -RT ln K

      -3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})

                x = 0.1417

Hence, at equilibrium

  •  [A] = 0.3 - 0.1417

       = 0.1583 M

  •  [B] = 0.4 - 0.1417

       = 0.2583 M

  •  [C] = 0.1417 M
5 0
3 years ago
A non metal which is good conductor of electricity
Neporo4naja [7]
The non metals good conductivity any to conduct electric city any time because the metals when we keep into any switch board while we touch we will get shock but non metals not like that
7 0
3 years ago
Read 2 more answers
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