As ________________ increases, air pressure decreases. A) pollution B) water vapor C) radiation D) altitude

If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?

photoshop1234 [79]

The combustion reaction is as expressed,

CxHy + O2 --> CO2 + H2O

The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.

Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.

moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14

The empirical formula for the hydrocarbon is therefore, CH₂.

7 0

2 years ago

In one to two sentences, explain a similarity and a difference between the particles in liquid water at 100ºC and the particles

mihalych1998 [28]

Answer

The particle theory is used to explain the properties of solids, liquids and gases. The strength of bonds (attractive forces) between particles is different in all three states.

6 0

3 years ago

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The Same force is applied to a 300 kg go kart and 100 kg wagon at the beginning of the race. Which time is true about their acce

ELEN [110]

The wagon will accelerate at a faster pace. Since the wagon is lighter than the go kart, the same force will have a stronger effect

5 0

2 years ago

A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°C. What is the magnitude of k at 95.0°C if Ea =

never [62]

Answer:

k ≈ 9,56x10³ s⁻¹

Explanation:

It is possible to solve this question using Arrhenius formula:

$ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )$

Where:

k1: 1,35x10² s⁻¹

T1: 25,0°C + 273,15 = 298,15K

Ea = 55,5 kJ/mol

R = 8,314472x10⁻³ kJ/molK

k2 : ???

T2: 95,0°C+ 273,15K = 368,15K

Solving:

$ln\frac{k2}{k1} = 4,257$

$\frac{k2}{k1} = 70,593$

${k2} = 9,53x10^3 s^{-1}$

I hope it helps!

5 0

2 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago