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2 years ago

9

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coeffici

ents are:
ammonia (g) +
oxygen (g)
nitrogen monoxide (g) +
water (g)

1 answer:

tatuchka [14]2 years ago

5 0

The values of the coefficients would be 4, 5, 4, and 6 respectively.

<h3>Balancing chemical equations</h3>

The equation of the reaction can be represented by the following chemical equation:

ammonia (g) + oxygen (g) ---> nitrogen monoxide (g) + water (g)

$4NH_3(g)$ + $5O_2(g)$ ---> $4NO(g)$ + $6H_2O(g)$

Thus, the coefficient of ammonia will be 4, that of oxygen will be 5, that of nitrogen monoxide will be 4, and that of water will be 6.

More on balancing chemical equations can be found here: brainly.com/question/15052184

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Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

2 years ago

According to Wookieepedia, Kylo Ren stands 1.89 m tall and his mass is 89 kg. Calculate his earth-weight in lbm. (2 pts) a. b. A

harina [27]

Answer:

a) 6312.12 lbm

b) D = 11.843 inch

Explanation:

GIVEN DATA:

Height of person = 1.89 m = 74.41 inch

mass of person m = 89 kg = 196.211 lbm

a) person earth weight =mg

where g is acceleration due to gravity = 32.17 ft/sec^2

$= 196.211\times 32.17 = 6312.12 lbm$

b) waist to height ratio = 0.5

$waist\ to\ height\ ratio = \frac{circumference\ of\ the\ waist}{height\ of\ the\ person}$

circumference of waist = $height\ of\ person \times waist\ to\ height ratio$

$= 74.41 inch \times 0.5$

=37.21 inch

we know that circumference is given as $= 2\pi r\ or\ \pi D= 37.21$

D = 11.843 inch

6 0

3 years ago

Need help please omg Perform each of the following conversions being sure to set up the appropriate conversion factor in each ca

sertanlavr [38]

Answer:

The answer to your question is:

a) 31.75 cm

b) 0.475 miles

c) 2.44 yards

d) 11496.04 inches

Explanation:

Convert

a) 12.5 in to cm

1 in ------------------- 2.54 cm

12.5 in ---------------- x

x = 12.5(2.54)/1 = 31.75/ = 31.75 cm

b) 2513 ft to miles

1 mile -------------- 5280 ft

x miles ------------ 2513 ft

x = 2513(1)/5280 = 0.475 miles

c) 2.23 m to yards

1 m ------------- 1,094 yards

2.23 m ---------- x

x= 2.23x1.094/1 = 2.44 yards

d) 292 m to inches

1 m ---------------- 39.37 inches

292 m ------------- x

x = 292 x 39.37/1 = 11496.04 inches

6 0

3 years ago

Which is NOT a shape of epithelial cells?

spayn [35]

B.should be *columnar, but C.Circular is the answer

4 0

3 years ago