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4 years ago

When are larger atoms least likely to be reactive

Chemistry

1 answer:

Sever21 [200]4 years ago

5 0

Answer:

when they have eight valence electrons.

Explanation:

When atoms do not have a full valence shell, the atoms are more likely to react with other and vice versa.

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What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin

Contact [7]

Answer: The total volume of the gaseous products is 1044.29 L

Explanation:

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = $\frac{1}{22.4}\times 116=5.18mol$ of butane

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

For carbon dioxide:

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = $\frac{8}{2}\times 5.18=20.72mol$ of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

For water vapor:

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = $\frac{10}{2}\times 5.18=25.9mol$ of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

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3 years ago

5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure water?

Margaret [11]

Answer:

$\boxed {\boxed {\sf 2.8 \ m }}$

Explanation:

The formula for molality is:

$m=\frac{moles \ of \ solute}{kg \ of \ solvent}$

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

$moles= 0.210 \ mol \\kilograms = 0.075 \ kg$

Substitute the values into the formula.

$m= \frac{ 0.210 \ mol }{0.075 \ kg}$

Divide.

$m= 2.8 \ mol/kg= 2.8 \ m$

The molality is 2.8 moles per kilogram

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3 years ago

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Answer:

No.D is the molecules that has trigonal pyramidal sape

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4 years ago

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What volume in mt, of 0.5a M1HCI solution is needed to neutralize 77 ml of 1.54 M NaOH solution?

Rainbow [258]

Answer:

237.2 mL.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HCl = (XMV) NaOH.

For HCl; X = 1, M = 0.5 M, V = ??? mL.

For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.

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Why do doctors worry more about a fever of a couple of degrees Celsius than a fever of a couple of degrees Fahrenheit

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Because Celcius is a larger amount than Fahrenheit. The units are farther apart.

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4 years ago

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When are larger atoms least likely to be reactive￼

When are larger atoms least likely to be reactive