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spayn [35]
3 years ago
6

When the temperature of an object changes by 10 °C the same temperature change in Kelvins would be

Chemistry
1 answer:
kumpel [21]3 years ago
8 0
The answer to this question would be: <span> 10 °K

Kelvin and Celcius scales are different by 273</span> degrees but their ratio is the same. One degree in Kelvin is equal to one degree in Celcius. That mean,  10 °C change in Celcius would be same as <span> 10 °K changes in Kelvin too. </span>
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nekit [7.7K]

Answer:

I think it's

there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium

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3 years ago
A testable question is one that _____.
cestrela7 [59]

Answer:

One that “Can be answered by conducting an experiment”

Explanation:

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3 years ago
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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
frez [133]

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

4 0
3 years ago
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For neon, determine the moles and mass contained in a 29.9−L volume at a pressure of 481.1 torr at a temperature of 300.0 K.
kondor19780726 [428]

Answer:

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Explanation:

Use the Ideal Gas Law, PV = nRT.

Make sure to use the correct ideal gas constant R. You can either put R in torr, or you can change the pressure to atm. I've just used the torr ideal gas constant.

481.1 torr * 29.9 L = n 62.364 LTorr/molK * 300 K

14384.89 = 18709.2n

n = <u>.7689 mol</u>

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8 0
3 years ago
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Answer:  

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