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Shalnov [3]
3 years ago
6

The compound with the molecular formula C4H10 is called

Chemistry
2 answers:
Alexus [3.1K]3 years ago
5 0

Answer:

Butane

...............

mel-nik [20]3 years ago
4 0

Answer:

butane

Explanation:

I hope this helps!

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Element: calcium symbol: Ca Atomic weight: g Mass of one mole: g/mol
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153g/mols I think this is the answer but not 100% sure.


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How are evaporation and condensation related?
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Both evaporation and condensation are in the water cycle. evaporation is when water changes from liquid to gas meanwhile condensation is when water vapor cools and goes from gas to liquid.
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Combustion analysis of 1.200 g of an unknown compound containing carbon, hydrogen, and oxygen produced 2.086 g of CO2 and 1.134
timurjin [86]
<span>the empirical formula is C3H8O2 You need to determine the relative number of moles of hydrogen and carbon. So you first calculate the molar mass of CO2 and H20 Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 Now calculate the number of moles of CO2 and H2O you have Moles CO2 = 2.086 g / 44.0087 g/mole = 0.0474 mole Moles H2O = 1.134 g / 18.01488 g/mole = 0.062948 mole Calculate the number of moles of carbon and hydrogen you have. Since there's 1 carbon atom per CO2 molecule, the number of moles of carbon is the same as the number of moles of CO2. But since there's 2 hydrogen atoms per molecule of H2O, The number of moles of hydrogen is double the number of moles of H2O Moles Carbon = 0.0474 Moles Hydrogen = 0.062948 * 2 = 0.125896 Now we need to determine how much oxygen is in the compound. Just take the mass of the compound and subtract the mass of carbon and hydrogen. What's left will be the mass of oxygen. Then divide that mass by the atomic weight of oxygen to get the number of moles of oxygen we have. 1.200 - 0.0474 * 12.0107 - 0.125896 * 1.00794 = 0.503797 Moles oxygen = 0.503797 / 15.999 = 0.031489 So now we have a ratio of carbon:hydrogen:oxygen of 0.0474 : 0.125896 : 0.031489 We need to find a ratio of small integers that's close to that ratio. Start by dividing everything by 0.031489 (selected because it's the smallest value) getting 1.505288 : 3.998095 : 1 The 1 for oxygen and the 3.998095 for hydrogen look close enough. But the 1.505288 for carbon doesn't work. But it looks like if we double all the numbers, we'll get something close to an integer for everything. So do so. 3.010575 : 7.996189 : 2 Now this looks good. Rounding everything to an integer gives us 3 : 8 : 2 So the empirical formula is C3H8O2</span>
5 0
3 years ago
Which of these is an example of food engineering?
Andrej [43]

Answer:

a.labeling food with an ingredient list to inform consumers

Explanation:

3 0
3 years ago
Read 2 more answers
At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 (
Sveta_85 [38]

<u>Answer:</u> The equilibrium constant for the above reaction is 1.31

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 0.3411 moles

Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

                   N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>         0.3411      1.661

<u>At eqllm:</u>     0.3411-x  1.661-3x      2x

Evaluating for 'x', we get:

\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141

Volume of the container = 2.50 L

The expression of K_c for the above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

We are given:

[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M

[N_2]=\frac{0.2001}{2.5}=0.08004M

[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M

Putting values in above expression, we get:

K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31

Hence, the equilibrium constant for the above reaction is 1.31

5 0
3 years ago
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