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Novosadov [1.4K]
2 years ago
10

Consider the equations below. (1) Fe2O3(s) → 2Fe(s) + 3 2 O2(g) (2) Fe2O3(s) + 3CO(g) → 2Fe(s)+3CO2(g) Which equation must be ad

ded to equation (1) to produce equation (2)? A. CO(g)
Chemistry
1 answer:
slega [8]2 years ago
7 0

Answer:

C. 3CO(g)+\frac{3}{2}O2(g)→3CO2(g)

Explanation:

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A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increa
MatroZZZ [7]

Answer:

T2= 7.3°C

Explanation:

To solve this problem we will use Charles law equation i.e,

V1/T1 = V2/T2    

Given data

V1 = 269.7 L

T1 = 6.12 °C

V2= 320.4 L

T2=?

Solution:

Now we will put the values in equation

269.7 L / 6.12°C  = 320.4 L / T2

T2=  320.4 L × 6.12°C/ 269.7 L

T2= 1960.85 °C. L /269.7 L

T2= 7.3°C

3 0
2 years ago
I need help pls help me I give brainless
Delicious77 [7]

Answer: the answer is nucleus duh!!

Explanation:

3 0
3 years ago
Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the
mojhsa [17]

Answer:

Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the

temperature of the water. The water in the beaker began to boil when it reached the temperature of

100'C. If Mario completes the same experiment with 25mL of water, what would happen to the boiling

point?

a) The water will not reach a boil.

b) The boiling point of water will increase.

c) The boiling point of water will decrease.

d) The boiling point of water will stay the same.

Explanation:

6 0
3 years ago
For the reaction, Cl2 + 2KBr --> 2KCl + Br2, how many moles of potassium chloride, KCl, are produced from 102g of potassium b
Sedaia [141]

Solution :

From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .

Moles of KBr in 102 g of potassium bromide.

n = 102/119.002

n = 0.86 mole.

So, number of miles of KCl produced are also 0.86 mole.

Mass of KCl produced :

m = 0.86 \times Molar \  mass \ of \  KCl\\\\m = 0.86 \times 74.5 \ gram \\\\m = 64.07\  gram

Hence, this is the required solution.

5 0
2 years ago
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
Natasha_Volkova [10]

Answer: 0.52V

Explanation:

Ecell = Ecell(standard) - [(0.0592 logQ)/n]

Q = product of the quotient

n = no of electrons transferred = 2

Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]

Ecell = 0.63 - 0.0194

Ecell = 0.5205V

5 0
3 years ago
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