All categories

3 years ago

How many moles of KI are present in 0.85L of a 0.55 M KI solution?

Chemistry

1 answer:

Mekhanik [1.2K]3 years ago

6 0

Answer: Amount of moles is 0.47 mol

Explanation: Amount of moles n = cV = 0.55 M · 0.85 l = 0.4675 mol

You might be interested in

Consider the following balanced equation:3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)If 24.2 moles of Na3PO4(aq) reac

Tomtit [17]

Answer:

69.7% is percent yield

Explanation:

Based on the reaction:

3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)

2 moles of Na3PO4 react producing 6 moles of NaNO3.

As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:

24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =

72.6 moles of NaNO3

As there are produced 50.6 moles of NaNO3, percent yield is:

50.6 moles NaNO3 / 72.6 moles NaNO3 =

<h3>69.7% is percent yield</h3>

7 0

3 years ago

Can the groups of the periodic table be numbered in two different ways

Vadim26 [7]

The groups (left to right) are numbered from 1 to 18. they arent numbered any other way.

period (up to down.) are numbered from 1-7

3 0

4 years ago

Convert 1.00 us gallon in kilometer (kL)

Murljashka [212]

The answer is 4.54km

5 0

2 years ago

What will happen as the South Pole magnet is moved closer and closer to the South Pole of another magnet

mart [117]

Answer:it will be malfunction

Explanation:

5 0

4 years ago

Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO

zavuch27 [327]

Answer:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

$2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons 4NOBr ( g )$

The suitable equilibrium constant turns out:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Or in terms of the initial equilibrium constant:

$K_2=K_1^2$

Since the second reaction is a doubled version of the first one.

Best regards.

5 0

3 years ago