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Mazyrski [523]
3 years ago
14

What is the percent yield of carbon dioxide of 2.5 mole of oxygen react and 32.4 g of carbon dioxide are produced?

Chemistry
1 answer:
Ugo [173]3 years ago
6 0

Answer:

86

Explanation:

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24. The air in a small 250 cm birthday balloon is at a pressure of 760 torr. A boy sits on it at
kvv77 [185]

Answer:

1.14atm

Explanation:

Given parameters:

V1 = 250cm³ ;

              1000cm³ = 1dm³;  so this is 0.25dm³

P1  = 760torr

            760torr  = 1atm

       

V2  = 220cm³ ; 0.22dm³

Unknown:

New pressure = ?

Solution:

To solve this problem, we apply Boyle's law and we use the expression below:

       P1 V1 = P2V2

The unknown is P2;

            1 x 0.25  = P2 x 0.22

                 P2  = 1.14atm

7 0
3 years ago
Am I weird or is no one seeing my qn<br>free point by the way
Julli [10]

Answer:

 maybe they don't want to answer it  only if you give them a lot of points and brainliest they will answer it i think

Explanation:

5 0
2 years ago
Read 2 more answers
Calculate the mass m of 6.50 moles n of kbr m
crimeas [40]

Answer:

773.51495 grams

Explanation:

1 moles KBr to grams = 119.0023 grams

6.5*119.0023 = 773.51495 grams

8 0
3 years ago
Dalton’s Law CalculationA mixture of H₂, N₂ and Ar gases is present in a steel cylinder. The total pressure within the cylinder
Zolol [24]

Answer:

A) The partial presssure of CO₂ is 167 mm Hg

B) The partial presssure of N₂ is 354 mm Hg

C) The partial presssure of Ar is 235 mm Hg

D) The partial presssure of H₂ is 86 mm Hg

Explanation:

Dalton's law of partial pressures is basically expressed by the following statement:

The total pressure of a mixture is equal to the sum of the partial pressures of its components.

So initially we have:

P_{T}= total presure of the system (675 mm Hg).

P_{N_2}= partial pressure of N₂ (354 mm Hg).

P_{Ar}= partial pressure of Ar (235 mm Hg).

Using Dalton's law we can find the partial pressure of H₂:

P_{T}=P_{N_2}+P_{Ar}+P_{H_2}

675 mm Hg=354 mm Hg + 235 mm Hg + P_{H_2}

P_{H_2}= 675 mm Hg - 354 mm Hg - 235 mm Hg

P_{H_2}=86 mm Hg

If CO₂ gas is added to the mixture, at constant temperature, and the volume is the same, the difference between the new total pressure and the previous total pressure is equal to the partial pressure of CO₂.

P_{T}=P_{N_2}+P_{Ar}+P_{H_2}+P_{CO_2}

842 mm Hg= 354 mm Hg + 235 mm Hg + 86 mm Hg + P_{CO_2}

P_{CO_2}= 842 mm Hg - 354 mm Hg - 235 mm Hg - 86 mm Hg

P_{CO_2}= 167 mm Hg

4 0
3 years ago
A metal pellet with a mass of 100.0 g, originally at 116°C, is dropped into a cup of water, initially at
Alina [70]

Answer:

C, 42g

Explanation:

In thermal equilibrium, both bodies (metal pellet and water) both have the same final temperature (46.3°C).

Assuming no heat is lost to surroundings,

the energy lost from metal pellet = energy gained for water

Since E = mc∆T

(energy = mass x specific heat capacity x temperature change)

mc∆T (metal pellet) = mc∆T (water)

100 x 0.568 x (116-46.3) = m 4.184 (46.3 - 23.8)

3958.96 = 94.14m

m = 42g

6 0
3 years ago
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