All categories

3 years ago

Balancing Equations: _(NH4)3PO4 + _Pb(NO3)4 = _Pb3(PO4)4 + _NH4NO3 With work please.

Chemistry

1 answer:

Svetlanka [38]3 years ago

8 0

1)4
2)3
3)0
4)12
4(NH4)3 + 3Pb(NO3)4= Pb3(PO4)4 +12NH4NO3

You might be interested in

Example: Make 100 mL 0.05 M NaOH from a 1.5 M solution.<br>0.05 * 100 = 1.5 * ?

____ [38]

Answer:

3.33

Explanation:

M= 0.05*100/1.5= 3.33

4 0

3 years ago

What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m

r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.

4 0

3 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

meriva

Answer:

The standard enthalpy change for the reaction at $25^{0}\textrm{C}$ is -2043.999kJ

Explanation:

Standard enthalpy change ( $\Delta H_{rxn}^{0}$ ) for the given reaction is expressed as:

$\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$