All categories

3 years ago

how many grams of silver chloride (AgCI) can be produced if you start with 4.62 grams of barium chloride (BaCI2)

Chemistry

2 answers:

UkoKoshka [18]3 years ago

6 0

The chemical equation representing the reaction of silver nitrate with barium chloride:

$2AgNO_{3}(aq) + BaCl_{2}(aq)--> 2AgCl (s) + Ba(NO_{3})_{2}(aq)$

Given mass of barium chloride = 4.62 g

Moles of $BaCl_{2} = 4.62 g BaCl_{2}*\frac{1 mol BaCl_{2}}{208.23 g BaCl_{2}} = 0.0222 mol BaCl_{2}$

Moles of AgCl = $0.0222 mol BaCl_{2} * \frac{2 mol AgCl}{1 mol BaCl_{2}}$ = 0.0444 mol AgCl

Mass of AgCl = $0.0444 mol AgCl * \frac{143.32 g AgCl}{1 mol AgCl} = 6.36 g AgCl$

Molodets [167]3 years ago

3 0

The answer to your question would be 6.27 grams

HOPE THIS HELPS

PLZ MARK AS BRAINLIEST

You might be interested in

A 8L sample of gas at 250 K is cooled to 75 K. What is the volume of the gas after it is cooled?

Stells [14]

$by \: Charles'\: law: \: \\ \frac{v_{1}}{t_{1}} = \frac{v_{2}}{t_{2}} \\$

Where v is the volume(in L) and t is the temperature(in °K)

$\frac{8}{250} = \frac{v_{2}}{75} \\\\4/5 =v_2/3 \\\\\boxed{v_{2} = \frac{12}{5}}\\\\\\\huge{=2.5 l}$

7 0

2 years ago

Lina has worked hard to gather data for her experiments. With her last set of data she received, she is ready to publish her pap

nika2105 [10]

43.8 has 3 significant figures and 1 decimal.

<h3 /><h3>What are significant figures?</h3>

The term significant figures refer to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation.

All zeros that occur between any two non-zero digits are significant. For example, 108.0097 contains seven significant digits. All zeros that are on the right of a decimal point and also to the left of a non-zero digit are never significant. For example, 0.00798 contained three significant digits.

Hence, 43.8 has 3 significant figures and 1 decimal.

Learn more about significant figures here:

brainly.com/question/14359464

#SPJ1

8 0

2 years ago

What is the first step of a ladybug during the growth and development process

Sunny_sXe [5.5K]

Answer:

Damian here! (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧

The newly hatched larva is in its first instar, a developmental stage that occurs between molts. It feeds until it grows too big for its cuticle, or soft shell, and then it molts. After molting, the larva is in the second instar. Ladybug larvae usually molt through four instars, or larval stages, before preparing to pupate.

Explanation:

hope this helps? :))

7 0

4 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

3 years ago

An analytical chemist weighs out 0.055g of an unknown triprotic acid into a 250mL volumetric flask and dilutes to the mark with

Katarina [22]

Answer:

Mass of the unknown acid is 4.0g

Explanation:

The determine the molar mass of the unknown acid, the steps below can be followed

Firstly, determine the concentration of the acid, the formula below can be used;

ConcA × Va/ConcB × Vb = Na/Nb

Where ConcA is the concentration of the unknown acid

ConcB is the concentration of the NaOH base

Va is the volume of acid and Vb is the volume of base

Since, the titration was said to have reached an equivalent point, it means the number of moles of the acid (Na) was equal to the number of moles of the base (Nb) and thus both will be assumed to be 1

Thus

ConcA × 250/0.13 × 6.6 = 1/1

ConcA = 0.13 × 6.6/250

ConcA = 0.003432M

Then, determine the actual number of moles (n) of the unknown acid used,

ConcA = no of moles of acid/volume of acid (in dm³ or L)

To convert mL to L, we divide by 1000

Hence, 250ml = 0.25L

0.003432 = n/0.25

n = 0.003432 × 0.25

n = 0.01373 moles

To determine the molar mass;

n = mass/molar mass

The mass was given in the question to be 0.055g

Thus

0.01373 = 0.055/molar mass

molar mass = 0.055/0.01373

molar mass = 4.0g

5 0

3 years ago