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3 years ago

2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

Chemistry

1 answer:

Harrizon [31]3 years ago

4 0

It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.

Explanation:

2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂

We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

$100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}$

= 337.5 g AgCl

In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.

It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.

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Paul [167]

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2 years ago

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Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

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3 years ago

What is the actual name of compound-s that is used to treat arthritis?

Rudiy27

Reichsteins substance if I remember correctly.

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3 years ago

How many moles of Na2SO4 can be made with 7.28 mol of H2SO4?

Slav-nsk [51]

Answer:

7.28 mol Na2SO4

Explanation:

Since it is already in moles, all we have to do is use a molar ratio

A molar ratio is the proportions of reactants and products using the balanced equation. When writing a mole ratio, the given information must cross out with the right thing.

7.28 mol H2SO4 * 1 mol Na2SO4/1 H2SO4 = 7.28 mol Na2SO4

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3 years ago

An interaction of systems allows for the body to warm itself when it is cold. Which processes help the body warm up when it is c

Dmitry_Shevchenko [17]

Answer:

shivering

hair on the body standing up

goosebump forming

Explanation:

The processes that help the body warm-up from the available options include the shivering of the body, formation of goosebumps on the skin, and the standing up of hairs on the body.

When the temperature of the body falls below the setpoint or the environment is cold, a homeostatic response is triggered and a signal is sent from the control center to the muscles of the body. The muscles start shaking in order to generate heat to raise the temperature of the body. At the same time, the tiny muscles at the base of the hairs on the skin contract and pull the hairs erect, causing goosebumps in the process.

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