A + c = r solve for a

What is the difference between a diatomic molecue and a normal molecue

Mekhanik [1.2K]

If a diatomic molecule consists of two atoms of the same element, such as hydrogen (H2) or oxygen (O2), then it is said to be homonuclear. Otherwise, if a diatomic molecule consists of two different atoms, such as carbon monoxide (CO) or nitric oxide (NO), the molecule is said to be heteronuclear.

7 0

3 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

Inside the retina of our eyes are receptors called rods and cones. Rods measure the brightness we see and cones determine _____.

JulijaS [17]

I believe the answer is C. 20 characters

3 0

4 years ago

1. What is the mass of 2.65×1023 atoms of lithium? grams

masya89 [10]

Answer: 3.05 grams

Explanation:

To find the mass in grams, we need to use Avogadro's number and the molar mass of lithium.

Avogadro's number: 6.022×10²³ atoms/mol

Molar mass: 6.94 g/mol

$2.65*10^2^3 atoms*\frac{mol}{6.022*10^2^3 atoms} *\frac{6.94g}{mol} =3.05 g$

5 0

3 years ago

How many grams of krypton gas will occupy 2.55 liters at 1.25 atm if it occupies 3.75 liters at 0.750 atm at 255 K

tatiyna

Answer:

There will be 94.7 grams of krypton gas

Explanation:

Step 1: Data given

at 255K and 0.750 atm the volume is 3.75 L

At 1.25 atm the volume is 2.55 L

Molar mass of krypton = 83.798 g/mol

Step 2: Calculate number of moles

p*V = n*R*T

with p = the pressure of the krypton gas = 0.750 atm

with V = the volume of the krypton gas = 3.75 L

with n = the number of moles = TO BE DETERMINED

with R = the gasconstant = 0.08206 L*atm/ mol * K

with T = the temperature = 255 K

n = R*T / p*V

n = (0.08206 * 255)/ (0.750 * 3.75)

n = 0.744 moles

Step 3: Calculate new number of moles

P1*V1 = P2*V2

P1*V1 = n1*R*T

P2*V2 = n2*R*T

(P1*V1)/R*T = (P2*V2)/n2*R*T

Since R and T do not change we can write as followed:

(P1*V1) = (P2*V2) /n2

n2 = (P2*V2) / (P1*V1)

n2 = (2.55 *1.25)/(0.75*3.75)

n2 = 1.13 moles

Step 4: Calculate mass of krypton gas

mass = Number of moles * Molar mass

mass of krypton gas = 1.13 moles * 83.798 g/mol = 94.69 grams ≈ 94.7 grams

There will be 94.7 grams of krypton gas

5 0

3 years ago