What coefficient of 02 should be added so the number of atoms of oxygen is conserved on both sides of the reaction equation

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

Step 1: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

Step 2: Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

Step 3: Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

3 years ago

The gas in a cylinder has a volume of 7 liters at a pressure of 107kPa. The pressure of the gas is increased to 208kPa. Assuming

VashaNatasha [74]

By Boyles Law (P1V1=P2V2), substituting values in and solving for V2, we find that the new volume is 3.6 L

6 0

3 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

3 years ago

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing

rodikova [14]

Answer:

0.225 mol = 0.23 mol to 2 significant figures

Explanation:

Calculate the moles of oxygen needed to produce 0.090 mol of water

The equation of the reaction is given as;

2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O

From the equation of the reaction;

5 mol of O2 produces 2 mol of H2O

x mol of O2 produces 0.090 mol of H2O

5 = 2

x = 0.090

x = 0.090 * 5 / 2

x = 0.225 mol

8 0

3 years ago

How many moles of oxygen must be placed

Annette [7]

0.24 moles of oxygen must be placed in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C.

The variables given are Pressure, volume and temperature.

Explanation:

Given:

P = 2 atm

V = 3 litres

T = 25 degrees or 298.15 K by using the formula 25 + 273.17 = K

R = 0.082057 L atm/ mole K

n (number of moles) = ?

The equation used is of Ideal Gas law:

PV = nRT

n = $\frac{PV}{RT}$

Putting the values given for oxygen gas in the Ideal gas equation, we get

n = $\frac{2 x 3}{0.082157 x 298.15}$

= 0.24

Thus, from the calculation using Ideal Gas law it is found that 0.24 moles of oxygen must be placed in a container.

Ideal gas law equation is used as it tells the relation between temperature, pressure and volume of the gas.

3 0

3 years ago

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