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2 years ago

Net ionic equation for Calcium hydroxide reacting with ammonium Chloride?

Chemistry

2 answers:

Anton [14]2 years ago

8 0

Explanation:

ca (OH)2 + 2NH4CL------> CACL2 + 2NH3 + 2H2O

VMariaS [17]2 years ago

3 0

Answer:

Ca(OH)₂ (aq) + 2NH₄Cl (aq) → CaCl₂ (aq) + 2NH₃ (g) + 2H₂O (l)

Ca²⁺ + 2OH⁻ + 2NH₄⁺ + 2Cl⁻ → Ca⁺² + 2Cl⁻ + 2NH₃ (g) + 2H₂O (l)

2OH⁻ + 2NH₄⁺ → 2NH₃ (g) + 2H₂O (l)

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I need help solving this!!!

iogann1982 [59]

Search up on safari the name of the worksheet and also put “answer key” and it should probably give you the answers there

4 0

3 years ago

How many moles of helium are needed to fill a balloon to a volume of 5.3 L at 22 ℃ and 632 mmHg?

son4ous [18]

Answer:

0.18 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (0.8316×5.3)/(0.083×295)

n = 0.18 moles

6 0

3 years ago

Identify the type of igneous rock that is formed when magma cools above Earth's surface?

mylen [45]

Answer: cools down and keeps going

Explanation:

5 0

3 years ago

If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a

strojnjashka [21]

Answer : The final temperature of the mixture is, $22.14^oC$

Explanation :

First we have to calculate the mass of ethanol and water.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g$

and,

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g$

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.42J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol = 35.5 g

$m_2$ = mass of water = 45.0 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $8.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get:

$35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.14^oC$

Therefore, the final temperature of the mixture is, $22.14^oC$

3 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

3 years ago