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3241004551 [841]
1 year ago
13

What is the mass of silver bromide that precipitates when 2.96 g of iron(iii) bromide is combined with excess silver nitrate?

Chemistry
1 answer:
KIM [24]1 year ago
7 0

<u>5.6400 </u>is the mass of silver bromide that precipitates when 2.96 g of iron(iii) bromide is combined with excess silver nitrate.

<h3>Difference between silver bromide and iron(iii) bromide</h3>

  • Silver bromide (AgBr) is a soft, pale-yellow, water-insoluble salt well known (along with other silver halides) for its unusual sensitivity to light. This property has allowed silver halides to become the basis of modern photographic materials. AgBr is widely used in photographic films and is believed by some to have been used for making the Shroud of Turin. The salt can be found naturally as the mineral bromargyrite.

  • Iron(III) bromide is the chemical compound with the formula FeBr3. Also known as ferric bromide, this red-brown odourless compound is used as a Lewis acid catalyst in the halogenation of aromatic compounds. It dissolves in water to give acidic solutions.

Learn more about Silver bromide

brainly.com/question/16958040

#SPJ4

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PLEASE HELP!<br><br> See picture
Darya [45]

Answer:

See Explanation

Explanation:

The equation of the reaction;

KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)

Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles

Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.

Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles

Since the reaction is 1:1, 0.45 moles of K2SO4 is produced

Hence K2SO4 is the limiting reactant.

Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g

So;

1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles

Mass of excess KOH = 0.09 moles * 56.1056 g/mol  = 5 g of excess KOH

5 0
3 years ago
Assuming the input of energy continues for another 5.0 seconds, where will the particle be?
konstantin123 [22]

The answer is A. Equilibrium

3 0
3 years ago
How many grams of hydrogen are produced if 30.0 g of zinc reacts?
alekssr [168]
<span>0.925 grams if using hydrochloric acid in the reaction. 0.462 grams if using sulfuric acid in the reaction. 0.000 grams if using nitric acid in the reaction. Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is: Zn + 2 HCl ==> ZnCl2 + H2 So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen. Atomic weight zinc = 65.38 Atomic weight hydrogen = 1.00794 Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So 0.458855919 * 2 * 1.00794 = 0.92499847 grams. Rounding to 3 significant figures gives 0.925 grams. To show the assumption of the acid used, the balanced equation for sulfuric acid would be Zn2 + H2SO4 ==> Zn(SO4)2 + H2 Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid). If nitric acid were used, the reaction is 4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O Which means that NO hydrogen gas is generated. The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.</span>
3 0
3 years ago
When excess potassium hydroxide and H2SeO3 react whats the net ionic reaction for the acid-base rxn
Bad White [126]

The required net ionic equation is; 2H^+(aq) + 2OH^-(aq)-----> 2H2O(l)

The molecular reaction equation is;

H2SeO3(aq) + 2KOH(aq) -----> K2SeO3(aq) + 2H2O(l)

The complete ionic equation is;

2H^+(aq) + SeO3^2-(aq) + 2K^+(aq) + 2OH^-(aq)-----> 2K^+(aq) + SeO3^2-(aq) + 2H2O(l)

Net ionic equation;

2H^+(aq) + 2OH^-(aq)-----> 2H2O(l)

We can clearly see that this is a neutralization reaction hence water is the product of the net ionic equation.

Learn more:brainly.com/question/25150590

4 0
2 years ago
A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of
jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is 1.6\times 10^2mL

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

Let us assume:

P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K

Putting values in above equation, we get:

\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL

Hence, the volume when the pressure and temperature has changed is 1.6\times 10^2mL

5 0
3 years ago
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