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zhenek [66]
3 years ago
15

How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)? 2AgNO3 + CaC

l2 → 2AgCl + Ca(NO3)2
A. 2.2 mol CaCl2

B. 3.1 mol CaCl2

C. 6.2 mol CaCl2

D. 12.4 mol CaCl2
Chemistry
1 answer:
nexus9112 [7]3 years ago
6 0

Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction

2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).

From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂

Henceforth, 6.2 moles of AgNO₃ reacts with \frac{6.2}{2} = 3.1 moles of CaCl₂.

1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-

A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.

C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.

D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃

Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.

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How to find the final temperature
sveta [45]

Answer:

The final temperature will be "12.37°".

Explanation:

The given values are:

mass,

m = 0.125 kg

Initial temperature,

c = 22.0°C

Time,

Δt = 4.5 min

As we know,

⇒  q=mc \Delta t

On putting the estimated values, we get

⇒     =0.125\times 22.0\times 4.5

⇒     =12.37^{\circ}

8 0
3 years ago
A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resu
Deffense [45]

Answer:

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown non-electrolyte = 2.17 grams

Mass of chloroform = 225.0 grams

The freezing point of the resulting solution is –64.2 °C

The freezing point of pure chloroform is – 63.5°C

kf = 4.68°C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ ΔT = The freezing point depression = T (pure solvent) − T(solution) = -63.5°C + 64.2 °C = 0.7 °C

⇒i = the van't Hoff factor = non-electrolyte = 1

⇒ kf = the freezing point depression constant = 4.68 °C/m

⇒ m = molality = moles unknown non-electrolyte / mass chloroform

0.7 °C = 1 * 4.68 °C/m * m

m = 0.150 molal

Step 3: Calculate moles unknown non-electrolyte

molality = moles unknown non-electrolyte / mass chloroform

Moles unknown non-electrolyte = 0.150 molal * 0.225 kg

Moles unknown non-electrolyte = 0.03375 moles

Step 4: Calculate molecular mass unknown non-electrolyte

Molar mass = mass / moles

Molar mass = 2.17 grams / 0.03375 moles

Molar mass = 64.3 g/mol

The molar mass of the unknown non-electrolyte is 64.3 g/mol

6 0
3 years ago
What is the mass of 7 x 10^28 atoms of Fe?
Nostrana [21]

Answer:

6 x 10⁶ g Fe

Explanation:

Step 1: Set up dimensional analysis

7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)

Step 2: Multiply, divide, and cancel out units

atoms Fe and atoms Fe cancel out.

mol Fe and mol Fe cancel out.

We should be left with g Fe.

7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe

116279(55.85) = 6.49 x 10⁶ g Fe

Step 3: Sig figs

There is only 1 sig fig in this problem.

6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe

4 0
3 years ago
What causes the force that pushes magma from the magma chamber through the pipe and vent of a volcano?
tatuchka [14]

the force that pushes magma from the magma chamber through the pipe and vent of a volcano is expanding gases.




Hope this helps you succeed. :D

4 0
3 years ago
Read 2 more answers
What mass of ice can be melted with the same quantity of heat as required to raise the temperature of 3.00 mol H2O(l) by 50.0°C?
lions [1.4K]

Answer:

m=33.9g

Explanation:

Hello,

In this case, we can first compute the heat required for such temperature increase, considering the molar heat capacity of water (75.38 J/mol°C):

Q=nCp \Delta T=3.00mol*75.38\frac{J}{mol\°C} *50.0\°C\\\\Q=11307J

Afterwards, the mass of ice that can be melted is computed by:

Q=n \Delta _{fus}H

So we solve for moles with the proper units handling:

n=\frac{Q}{\Delta _{fus}H} =\frac{11307J}{6010\frac{J}{mol} } =1.88mol

Finally, with the molar mass of water we compute the mass:

m=1.88mol*\frac{18g}{1mol}\\ \\m=33.9g

Best regards.

7 0
3 years ago
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