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timurjin [86]
3 years ago
14

The average temperature at the South Pole In January is - 35.4 °C.

Chemistry
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

-31.72°F

Explanation:

(-35.4°C × 9/5) + 32 = -31.72°F

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Which sample of matter is classified as a solution? 1. H2O(s) 2. H2O(l) 3. CO2(g) 4. CO2(aq)
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Answer : Option 4) CO_{2}_{(aq)}

Explanation : CO_{2}_{(aq)} is the only sample of matter which can be classified as a solution. As the solution can be defined as a liquid mixture which contains a minor component (the solute) that is uniformly distributed within the major component (the solvent).

In this case, the solute is CO_{2} which is dissolved in water which acts as an solvent. Also, it has a subscript which is aq. which means aqueous, is often given to the solution in which the solvent is water.

Therefore, CO_{2}_{(aq)} is the correct answer.

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes
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Answer : The mass of MnCO_3 required are, 35 kg

Explanation :

First we have to calculate the mass of MnO_2.

The first step balanced chemical reaction is:

2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2

Molar mass of MnCO_3 = 115 g/mole

Molar mass of MnO_2 = 87 g/mole

Let the mass of MnCO_3 be, 'x' grams.

From the balanced reaction, we conclude that

As, (2\times 115)g of MnCO_3 react to give (2\times 87)g of MnO_2

So, xg of MnCO_3 react to give \frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg of MnO_2

And as we are given that the yield produced from the first step is, 65 % that means,

60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg

The mass of MnO_2 obtained = 0.4542x g

Now we have to calculate the mass of Mn.

The second step balanced chemical reaction is:

3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3

Molar mass of MnO_2 = 87 g/mole

Molar mass of Mn = 55 g/mole

From the balanced reaction, we conclude that

As, (3\times 87)g of MnO_2 react to give (3\times 55)g of Mn

So, 0.4542xg of MnO_2 react to give \frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg of Mn

And as we are given that the yield produced from the second step is, 80 % that means,

80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg

The mass of Mn obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of MnCO_3 required are, 35 kg

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3 years ago
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