All categories

3 years ago

What is the formula for Tetrachlorine pentaiodide

Chemistry

1 answer:

rjkz [21]3 years ago

4 0

Answer:

subscribe to me on you-tube for brainliest custom link since u cant do a you-tube link :/

https://screenshare.host/B7N8NT

you-tube name janathion loran

Explanation:

You might be interested in

A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

laiz [17]

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 $= 10$ ml

Molar concentration of Ca(NO3)2 $= 0.001$

Volume of NaF $= 10$ ml

Molar concentration of NaF $= 0.0001$

Ksp for CaF2 $= 3.2 * 10^ {-11}$

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

$= 10 * 0.001\\= 0.01$

$[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}$

Moles of F- ion

$= 10 * 0.0001\\= 0.001$

$[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}$

$Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}$

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

5 0

3 years ago

A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g

svetoff [14.1K]

Answer:

The answer to your question is a) 51.07 g of NiBr₂ b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

mass of Br = 37.3 g

Balanced chemical reaction

Ni + Br₂ ⇒ NiBr₂

Process

1.- Find the atomic mass of the reactants and the molar mass of the product

Ni = 59 g

Br = 79.9 x 2 = 159.8 g

NiBr₂ = 59 + 159.8 = 218.8 g

2.- Find the limiting reactant

theoretical yield Ni/Br₂ = 59/159.8 = 0.369

experimental yield Ni/Br₂ = 67.8/37.3 = 1.81

The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

159.8 g of Br₂ --------------- 218.8 g of NiBr₂

37.3 g of Br₂ -------------- x

x = (37.3 x 218.8) / 159.8

x = 8161.24/159.8

x = 51.07 g of NiBr₂

4.- Find the excess reactant

The excess reactant is Nickel

59 g of Ni ---------------- 159.8 g of Br₂

x ---------------- 37.3 g of Br₂

x = (37.3 x 59)/159.8

x = 2200.7/159.8

x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

= 54 g

5 0

3 years ago

Will marks brainly if it isn’t a copied answer!!!

Neporo4naja [7]

What kind of control like scientific?

5 0

3 years ago

Read 2 more answers

What would happen if carbon -14 gained 1 proton and lost 1 neutron?

sleet_krkn [62]

D. To answer this question refer to the periodic table and think logically about it. Adding one proton increases the atomic number so you go along the row to Nitrogen. If you lose one neutron then the atomic mass decreases by 1 so 14-1 is 13.

4 0

3 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

3 years ago