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2 years ago

How do you convert kilograms to pounds?

1 answer:

4 0

Answer: (1 Kilogram = 2.20462 pounds) . There are 2.2046226218 lb in 1 kilogram. To convert kilograms to pounds, multiply your figure by 2.205 for an approximate result. 1 kilogram is also equal to 2 lb and 3.27396195 oz. Working out a rough estimate in your head for converting to pounds and ounces may be tricky - remember that there are 16 ounces in a pound.

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How many grams of sodium acetate are in solution in the third beaker?

Kipish [7]

Answer:

46g of sodium acetate.

Explanation:

The data is: <em>Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156g of the salt in 100 mL of water at 100°C and then slowly cooling it to 20°C. Because the solubility of sodium acetate in water at 20°C is 46g per 100mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.</em>

The third solution is the result of the equilibrium in the solution at 20°C. As the maximum quantity that water can dissolve of sodium acetate at this temperature is 46g per 100mL and the solution has 100mL <em>there are 46g of sodium acetate in solution. </em>The other sodium acetate precipitate because of decreasing of temperature.

I hope it helps!

6 0

3 years ago

Examples of noble gases

LekaFEV [45]

Natural gas, desolate

4 0

3 years ago

Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium

Eddi Din [679]

Explanation:

It is known that $pK_{a}$ value of acetic acid is 4.74. And, relation between pH and $pK_{a}$ is as follows.

pH = pK_{a} + log $\frac{[CH_{3}COOH]}{[CH_{3}COONa]}$

= 4.74 + log $\frac{1.00}{1.00}$

So, number of moles of NaOH = Volume × Molarity

= 71.0 ml × 0.760 M

= 0.05396 mol

Also, moles of $CH_{3}COOH$ = moles of $CH_{3}COONa$

= Molarity × Volume

= 1.00 M × 1.00 L

= 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

$CH_{3}COONa + NaOH \rightarrow CH_{3}COOH$

Initial : 1.00 mol 1.00 mol

NaoH addition: 0.05396 mol

Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)

= 0.94604 mol = 1.05396 mol

As, pH = pK_{a} + log $\frac{[CH_{3}COONa]}{[CH_{3}COOH]}$

= 4.74 + log $\frac{0.94604}{1.05396}$

= 4.69

Therefore, change in pH will be calculated as follows.

pH = 4.74 - 4.69

= 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

8 0

3 years ago

A student wants to compare how easily a metal and a plastic spoon of the same temperature can transfer thermal energy. Which of

Artist 52 [7]

Measure the effort required to bend each spoon.

5 0

3 years ago

Read 2 more answers

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