All categories

3 years ago

The equation NaCI + AgF -> NaF + AgCI is an example of what type of reaction

Chemistry

1 answer:

PilotLPTM [1.2K]3 years ago

3 0

Double replacement. The metal ions have exchanged partners.

“Single replacement” is incorrect. In single displacement, one element displaces another element in a compound.

“Decomposition” is incorrect. In decomposition, a single substance changes into two or more substances.

“Combustion” is incorrect because it usually involves a reaction with oxygen .

You might be interested in

1. How many moles of LiBr are present in 100 mL of 1.25M LIBr solution?

Svet_ta [14]

Answer:

https://www.webassign.net/question_assets/wertzcams3/appendix.pdf

7 0

3 years ago

A substance that dissolves another substance is a _____________________

Scorpion4ik [409]

Answer:

b it is easy its b

Explanation:

7 0

3 years ago

Read 2 more answers

If an element only partially conducts electricity, it would be part of the _______ category. *

dmitriy555 [2]

It would be in the transition metals

5 0

3 years ago

20.00 g of aluminum (al) reacts with 78.78 grams of molecular chlorine (cl2) all of each reactant is completely consumed and a s

marysya [2.9K]

The mass of the product is 98.78 g.

The word equation is

aluminum + chlorine → product

20.00 g + 98.78 g → x g

If each reactant is completely consumed, the Law of conservation of Mass tells us the mass of the product must be 98.78 g.

8 0

3 years ago

When 0.49 g of a molecular compound was dissolved in 20.00 g of cyclohexane, the freezing point of the solution was lowered by 3

IRISSAK [1]

Answer: The molecular mass of this compound is 131 g/mol

Explanation:

Depression in freezing point:

$\Delta T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$\Delta T_f$ = depression in freezing point = $3.9^oC$

$k_f$ = freezing point constant = $20.8^0C/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 0.49 g

$w_1$ = mass of solvent (cyclohexane) = 20.00 g

$M_2$ = molar mass of solute = ?

Now put all the given values in the above formula, we get:

$(3.9)^oC=1\times (20.8^oC/m)\times \frac{(0.49g)\times 1000}{M_2\times (20.00g)}$

$M_2=131g/mol$

Therefore, the molar mass of solute is 131 g/mol

7 0

3 years ago