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jeka94
3 years ago
7

Need help with balancing equations!

Chemistry
2 answers:
kirill [66]3 years ago
8 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the balanced chemical equation for given reaction ~

\sf Fe_2O_3 (s) +3 \:  CO (g) =2 \:  Fe(l) + 3\: CO_2 (g)

As we can see, the Coefficient of Fe (l) is 2

shusha [124]3 years ago
4 0

Answer:

Fe2O3 + 3CO → 2Fe + 3CO2

Explanation:

the numbers in front are the numbers you need

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Nimfa-mama [501]

Answer:

The hydrogen ion concentration in a solution, [H+], in mol L-1, can be calculated if the pH of the solution is known.

pH is defined as the negative logarithm (to base 10) of the hydrogen ion concentration in mol L-1 pH = -log10[H+] ...

[H+] in mol L-1 can be calculated using the equation (formula): [H+] = 10-pH

8 0
2 years ago
Earth is currently 150 million km away from the Sun. If Earth was 200 million km away from the Sun, why would it no longer be ab
timurjin [86]

Answer:

The temperature would be too cold

Explanation:

4 0
3 years ago
Manganese-58 has a half-life of about 3 seconds. If you have a 90.0 gram sample, how
Rus_ich [418]

Answer:

18.018 seconds.

Explanation:

Given that the half life of Manganese, Mn = 3 seconds. The initial sample mass = 90.0 gram, the final sample mass = 1.40 gram.

The general idea to the question is to look for the time it will take to decay from the initial mass that is 90 gram to 1.40 gram.

Therefore, we will be making use of the formula below;

J(t) = J(o) × (1/2)^t/t(hL).

Where t(hL) is the half life, t is the time taken, J(t)= mass after time,t and J(o) is the initial mass. So, let us slot in the values into the equation above.

1.4 = 90 × (1/2)^ t/3.

1.4/90 = (1/2)^t/3.

t/3 = log(0.5) (1.4/90).

+Please note that the 0.5 of the log is at the subscript).

That is the base 0.5 logarithm of (1.4/90) 0.01556 is 6.0060141295.

t = 3 × 6.0060141295.

t = 18.018 seconds.

4 0
3 years ago
A chemist heats the block of copper as shown in the interactive, then places the metal sample in a cup of oil at 25.00 °C instea
Mazyrski [523]

When the oil is added to the heated copper, the energy in the system is

conserved.

  • The mass of the oil in the cup, is approximately <u>64.73 grams</u>.

Reasons:

The question parameters are;

Temperature of the oil in the cup = 25.00°C

Final temperature of the oil and copper, T₂ = 27.33 °C

Specific heat of copper, c₂ = 0.387 J/(g·°C)

Specific heat capacity of oil, c₁ = 1.74 J/(g·°C)

Required:

The<em> mass of oil</em> in the cup.

Solution:

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating, T₂ = 65.17°C

Temperature of the copper after being placed in the cup of oil, T₂ = 27.33°C

Heat lost by copper = Heat gained by the oil

  • m₂·c₂·(T₂ - T₃) = m₁·c₁·(T₃ - T₁)

Therefore, we get;

17.920 × 0.387 × (65.17 - 27.33) = m₁ × 1.74 × (27.33 - 25)

262.4219136 = 4.0542·m₁

m₁ ≈ 64.73

  • The mass of the oil in the cup, m₁ ≈ <u>64.73 g</u>

Learn more here:

brainly.com/question/21406849

<em>Possible part of the question obtained from a similar question online, are;</em>

<em>The mass of the copper, m₂ = 17.920 g</em>

<em>Temperature of copper after heating = 65.17°C</em>

6 0
3 years ago
Determination of the chemical composition of the atmospheres of the planets is carried out most effectively by what type of stud
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ATMOSPHERIC CHEMISTRY
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