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3 years ago

How many molecules of butane (C4H10) are contained in a 9.213 g sample? How many hydrogen atoms are contained in this

Chemistry

1 answer:

Anit [1.1K]3 years ago

3 0

9.03 X $10^{22}$ molecules of butane is present in 9.213 grams of the sample.

1.5 atoms of hydrogen are present in the sample 9.123 grams of butane.

Explanation:

Data given:

mass of butane C4H10= 9.123 grams

atomic mass of butane = 58.12 grams/mole

number of molecules =?

number of hydrogen atoms in the sample=?

number of moles is calculated as :

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values in the above equation:

number of moles of butane = $\frac{9.213}{58.12}$

= 0.15 moles

number of molecules = number of moles x 6.02 x $10^{23}$ (Avagadro number)

Putting the values in above formula:

number of molecules of Butane = 0.15 X 6.023 X $10^{23}$

= 9.03 X $10^{22}$ molecules

Number of hydrogen atoms:

1 mole of C4H10 contains 10 atoms of H

0.15 moles of C4H10 will have x moles

$\frac{10}{1}$ = $\frac{x}{0.15}$

x = 0.15 atoms of hydrogen

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Answer:

1.60x10⁶ billions of g of CO₂

Explanation:

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1 mol of C₆H₁₂O₆ -------------------------- 6 moles of CO₂

Transforming for mass multiplying the number of moles by the molar mass:

180.156 g of C₆H₁₂O₆ ----------------- 264.06 g of CO₂

4.59x10² g ---------------- x

By a simple direct three rule:

180.156x = 121203.54

x = 672.77 g of CO₂ per day per human

So, in a year, 6.50 billion of human produce:

672.77 * 365 * 6.50 billion = 1.60x10⁶ billions of g of CO₂

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3 years ago

What happens when acetylene reacts with silver powder?

horrorfan [7]

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3 years ago

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

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4 years ago

The pka of hf is 3.2 determine the pkb of hf?

Julli [10]

Well, first we must remember that

$pK_{a}+pK_{b}=14$

This is because

$K_{a}*K_{b}=10^{-14}$

$-log(K_{a}*K_{b})=-log(10^{-14})\\-logK_{a}+-logK_{b}=-log(10^{-14})\\pK_{a}+pK_{b}=14$

So then

$pK_{b}=14-pK_{a}=14-3.2=1.8$

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4 years ago

Read 2 more answers

Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide tetrahydrate in enough water to make 450 mL

Vladimir [108]

Answer:

The molarity of the CoBr2•4H2O solution is 7.64 × 10-2 M

Explanation:

Cobalt (II) bromide tetrahydrate

• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+

• Bromide - anion from group 7A → -1 charge → symbol: Br-

• Tetrahydrate- tetra- means 4 and hydrate is H2O

The chemical formula of the compound is: CoBr2•4H2O

We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.

We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.

•The molar mass of CoBr2•4H2O is:

CoBr2•4H2O

1 Co x 58.93 g/mol Co = 58.93 g/mol

2 Br x 79.90 g/mol Br = 159.80 g/mol

8 H x 1.008 g/mol H = 8.064 g/mol

4 O x 16.00 g/mol O = 64.00 g/mol

________________________________________

Sum = 290.79 g/ mo

The moles of CoBr2•4H2O is:

= 10.0 g CoBr2•4H2O x $\frac{ 1 mol CoBr_2 . 4H_2O}{290.79 g CoBr_2 . 4H_2O}$

= 0.0344 mol CoBr2•4H

We know that the volume of the solution is 450 mL.

We can now calculate for molarity:

Convert mL to L → 1 mL = 10-3 L

Formula:

Molarity (M)= Mole of solute / Liters of solution

= 0.0344 mol CoBr2•4H / 450 mL x 1 ml / 10^ -3 L

= 0.0764

= 7.64 × 10-2 mol/L

8 0

3 years ago