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Vaselesa [24]
3 years ago
9

Which of the following pairs of compounds will react to form a precipitate? Why?

Chemistry
1 answer:
Ivanshal [37]3 years ago
4 0

Answer:

calcium chloride and silver nitrate

Explanation:

A precipitate is often formed in a double replacement reaction when the reaction yields an insoluble product.

Consider the reaction between calcium chloride and silver nitrate;

CaCl2(aq) + 2AgNO3(aq) -------->2AgCl(s) + Ca(NO3)2(aq)

AgCl is insoluble in water, hence the reaction leads to the formation of a precipitate.

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Calcium + zinc nitrate ------->
Snezhnost [94]

Answer:

c

Explanation:

there will be displacement reaction taking place

8 0
3 years ago
Read 2 more answers
How many grams of sodium chloride are present in a 0.75 M solution with a volume of 500.0 milliliters?
saveliy_v [14]
The molarity of a solution is a type of expression of concentration equal to the number of moles solute per liter solution. In this problem, we are given the molarity equal to 0.75 M and a volume equal to 500 milliliters. <span>500 milliliters is equal to 0.5 liters. we multiply M and L to get the number of moles then multiply by the molar mass of NaCl. The answer is 21.92 grams.</span>
6 0
3 years ago
Molar mass is measured in units of
blsea [12.9K]
D. grams/ mole.
........
5 0
3 years ago
If I have 3.5 moles of C, and excess Fe2O3 , how many moles of Fe can I produce?
Wittaler [7]

Answer:

3.5 moles Fe

Explanation:

From the equation, Reaction of 2 moles of Fe₂O₃ with  1 mole of C produces 1 mole of Fe. When excess Fe₂O₃ is used, the only liming factor is C.

The ratio of amount of C used to the amount of Fe produced is 1:1

Therefore, if 3.5 moles of C are used,  3.5 moles of Fe are also produced.

6 0
3 years ago
Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many
Flauer [41]

Answer : The amount of formaldehyde permissible are, 5.4\times 10^{-6}g

Explanation : Given,

Density of air = 1.2kg/m^3=1.2g/L     (1kg/m^3=1g/L)

First we have to calculate the mass of air.

\text{Mass of air}=\text{Density of air}\times \text{Volume of air}

\text{Mass of air}=1.2g/L\times 6.0L

\text{Mass of air}=7.2g

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 5.4\times 10^{-6}g

Thus, the amount of formaldehyde permissible are, 5.4\times 10^{-6}g

8 0
3 years ago
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