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3 years ago

How do you remove unine odor from bedding?

1 answer:

3 0

You could try a chemical from the store that is made to take away odors, or you could try and use fabric softener to make it smell better:)

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Question 10 of 10

Artist 52 [7]

Answer:

d 1 1/2 oz.. of 80-proof liquor

7 0

3 years ago

The combustion of 1.685 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00 K to 302.16 K. The he

AfilCa [17]

Answer:

ΔH = - 2020.57 kJ/mol

Explanation:

Given that :

mass of propanol = 1.685 g

the molar molar mass = 60 g/mol

Thus; the number of moles = mass/molar mass

= 1.685 g/60 g/mol

= 0.028 g/mol

However ;

ΔH = heat capacity C × Δ T

Given that:

The temperature increases from 298.00 K to 302.16 K.

Then ;

Δ T = 302.16 K - 298.00 K

Δ T = 4.16 K

heat capacity C = 13.60 kJ/K

∴

ΔH = 13.60 kJ/K × 4.16 K

ΔH = 56.576 kJ

The equation of the given reaction can be represented as :

$C_3H_7OH_{(l)}+\dfrac{3}{2}O_{2(g)} \to 3CO_{2(g)} +4H_2O_{(l)}$

Thus for 0.028 mol of heat liberated; ΔH = 56.576 kJ

For 1 mole of heat liberated now:

ΔH = 56.576 kJ/0.028 mol

ΔH = 2020.57 kJ/mol

SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;

ΔH = - 2020.57 kJ/mol

5 0

3 years ago

Read 2 more answers

How many moles of sodium hydroxide are there in 1.0mL of 2.0M NaOH

kvasek [131]

To find the moles, you can use the following formula

moles= Molarity x Liters

Molarity= 2.0 M
Liters= 0.0010 Liters ---------------->>>>>>>>>> 1.0 mL= 0.0010 Liters

moles= 2.0 M x 0.0010 Liters= 0.0020 moles

4 0

3 years ago

Dolomite is a mixed carbonate of calcium and magnesium. Calcium and magnesium carbonates both decompose upon heating to produce

Setler79 [48]

Answer:

72.03 %

Explanation:

Total mass of dolomite = 9.66 g

Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol$

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

$Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol$

According to the reaction shown below:-

$MgCO_3\rightarrow MgO+CO_2$

$CaCO_3\rightarrow CaO+CO_2$

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = $\frac{x}{84.3139}\ mol$

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = $\frac{x}{84.3139}\times 40.3044 \ g$

Moles of CaO = $\frac{9.66 - x}{100.0869}\ mol$

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = $\frac{9.66 - x}{100.0869}\times 56.0774 \ g$

Given that total mass of the oxide = 4.84 g

Thus,

$\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84$

$\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84$

$-694.1618435x+45673.48749\dots =40843.38968\dots$

$x=\frac{4830.09780\dots }{694.1618435}$

$x=6.9582$

Thus, the mass of Magnesium carbonate = 6.9582 g

$\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100$

$\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%$

3 0

3 years ago

∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821

Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

6 0

3 years ago