Oxidation number of Mg3N2

5) Determine the mass if lithium hydroxide that is produced when 12.87 g of lithium

netineya [11]

Answer: 26.54 grams

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of lithium nitride}=\frac{12.87g}{34.83g/mol}=0.369moles$

$Li_3N+3H_2O\rightarrow 3LiOH+NH_3$

$Li_3N$ is the limiting reagent as it limits the formation of product and $H_2O$ is the excess reagent

According to stoichiometry :

As 1 moles of $Li_3N$ give = 3 moles of $LiOH$

Thus 0.369 moles of $O_2$ give = $\frac{3}{1}\times 0.369=1.108moles$ of $LiOH$

Mass of $LiOH=moles\times {\text {Molar mass}}=1.108moles\times 23.95g/mol=26.54g$

Thus 26.54 g of $LiOH$ will be produced from the given mass.

4 0

3 years ago

What is nacl monosaccharide disaccharide organic or inorganic?

bonufazy [111]

<span>inorganic Let's look at the choices and see why they work, or don't work. monosaccharide * Otherwise known as a simple sugar. And NaCl is definitely not a sugar of any type. So this is wrong. disaccharide * Complex sugar. And NaCl doesn't qualify either. organic * A definition of an organic compound is one that has carbon in it. NaCl has sodium and chlorine. No carbon at all, so this isn't the right answer. And I wish that organic was an earlier choice, since the sugars mentioned above are organic compounds. inorganic * This is the only possible choice. Salt is not an organic compound since it doesn't have carbon. So it can't be a sugar either. But it can and is inorganic.</span>

8 0

3 years ago

gold(iii) chloride reacts with Antimony(III) oxide and water to produce Gold Antimony(V) oxide and Hydrochloric acid balanced eq

zhuklara [117]

Balanced equation below

5 0

3 years ago

How many carbon atoms are there in .500 mol of CO2?

AURORKA [14]

Answer: There are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

Explanation:

According to the mole concept, there are $6.022 \times 10^{23}$ atoms present in 1 mole of a substance.

In a molecule of $CO_{2}$ there is only one carbon atom present. Therefore, number of carbon atoms present in 0.500 mol of $CO_{2}$ are as follows.

$1 \times 0.500 \times 6.022 \times 10^{23}\\= 3.011 \times 10^{23}$

Thus, we can conclude that there are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

6 0

2 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago