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3 years ago

CaCl2 Ca + CI2 Balance this equation

Chemistry

1 answer:

Elis [28]3 years ago

3 0

To balance Ca + Cl2 = CaCl2 you'll need to be sure to count all of atoms on each side of the chemical equation.

Once you know how many of each type of atom you can only change the coefficients (the numbers in front of atoms or compounds) to balance the equation for Calcium + Chlorine gas.

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6. A 50-year-old sample of Strontium-90 is found. The half-life of strontium is 29

Gnesinka [82]

Answer:

the sample has mass of 50mg initially

Explanation:

https://youtu.be/ietdukGMG5c

8 0

3 years ago

At 500 degree C, F_2 gas is stable and does not dissociate, but at 840 degree C, some dissociation occurs: F_2 (g) 2 F(g). A fla

bazaltina [42]

Answer:

2.73 is the equilibrium constant for the dissociation of $F_2$ gas at 840 degree Celsius.

Explanation:

$F_2(g)\rightleftharpoons 2F(g)$

Initial

0.600 atm 0

Equilibrium

(0.600 atm - p) 2p

Total pressure at equilibrium = P = 0.984 atm

P= 0.600 atm - p)+2p=0.984 atm

p = 0.384 atm

Partial pressure of the $F_2$ gas , $p_{f_2}$ = (0.600 atm - 0.384 atm)=0.216 atm

Partial pressure of the $F$ gas, $p_{f}$ = 2(0.384 atm)=0.768 atm

$K_p=\frac{(p_{F})^2}{p_{F_2}}$

$K_p=\frac{(0.768 atm)^2}{0.216 atm}=2.73$

2.73 is the equilibrium constant for the dissociation of $F_2$ gas at 840 degree Celsius.

7 0

4 years ago

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

ella [17]

Answer :

The mass of excess mass of $Na_2CO_3$ , $AgNO_3,Ag_2CO_3\text{ and }NaNO_3$ are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of $Na_2CO_3$ = 4.25 g

Mass of $AgNO_3$ = 7.50 g

Molar mass of $Na_2CO_3$ = 106 g/mole

Molar mass of $AgNO_3$ = 170 g/mole

Molar mass of $Ag_2CO_3$ = 276 g/mole

Molar mass of $NaNO_3$ = 85 g/mole

First we have to calculate the moles of $Na_2CO_3$ and $AgNO_3$ .

$\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 2 moles of $AgNO_3$ react with 1 mole of $Na_2CO_3$

So, 0.044 moles of $AgNO_3$ react with $\frac{0.044}{2}=0.022$ moles of $Na_2CO_3$

From this we conclude that, $Na_2CO_3$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

The excess mole of $Na_2CO_3$ = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of $Na_2CO_3$ .

$\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g$

Now we have to calculate the moles of $Ag_2CO_3$ .

As, 1 moles of $AgNO_3$ react to give 1 moles of $Ag_2CO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $Ag_2CO_3$

Now we have to calculate the mass of $AgCO_3$ .

$\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g$

Now we have to calculate the moles of $NaNO_3$ .

As, 2 moles of $AgNO_3$ react to give 2 moles of $NaNO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $NaNO_3$

Now we have to calculate the mass of $NaNO_3$ .

$\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g$

6 0

4 years ago

Give 4 examples of Uniformitarianism

ipn [44]

Answer:

Good examples are the reshaping of a coastline by a tsunami, deposition of mud by a flooding river, the devastation brought by a volcanic explosion, or a mass extinction caused by an asteroid impact.

6 0

3 years ago

Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 32.5°c.

Korvikt [17]

The rms speed of a gas can be calculated using the following rule:
Vrms = sqrt[(3RT) / M] where
R is the gas constant = 8.314 <span>J K^−1 mol^−1
T is the temperature = 32.5 + 273 = 305.5 degree kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens in the equation to get the rms speed as follows:
Vrms = sqrt [(3*8.314*305.5) / 0.028]
Vrms = 521.665 m/sec</span>

5 0

4 years ago