How many is too many

A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of wa

Rudik [331]

Answer:

1.43 (w/w %)

Explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

1 mole of HCl reacts per mole of ammonia.

Mass of NH3 is obtained as follows:

Moles HCl:

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = Moles NH3

Mass NH3 in the aliquot:

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

6 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

Calculate the number of O atoms in 0.150 g of CaSO4 · 2H2O.

natulia [17]

Answer:

There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.

Explanation:

Mass of calcium sulfate crystal = m = 0.150 g

Molar mass of calcium sulfate crystal = M = 172 g/mol

Moles of magnesium nitrate = n

$n=\frac{m}{M}$

$n=\frac{0.150 g}{172 g/mol}=0.0008721 mol$

1 mole of calcium sulfate crystal has 6 moles of oxygen atoms. Then 0.004446 moles calcium sulfate crystal will have :

$6\times 0.0008721 mol=0.0052326mol\approx 0.00523 mol$

There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.

8 0

3 years ago

Semi-conductors have properties of both metals and non metals.which are two examples of metalloids

balu736 [363]

The six metalloids are boron, silicon, germanium, arsenic, antimony, and tellerium.

7 0

3 years ago

The term ___________ refers to the grinding away of rock by rock particles carried by water, ice, wind, or gravity

eimsori [14]

Answer:

isn't it erosion

Explanation:

if i am wrong sorry have a good day

7 0

3 years ago