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2 years ago

4. If you put 1 mL of coffee into 9 mL of water, how much was the coffee diluted?

Chemistry

2 answers:

garri49 [273]2 years ago

7 0

1 in 10, or 1/10, or 0.1 dilution factor It may be wrong but this is what I have!

VARVARA [1.3K]2 years ago

7 0

Answer : The coffee diluted was, 1 in 10, or 1/10, or 0.1 dilution factor.

Explanation :

Dilution : It is defined as the solute concentration of a solution decreased by adding the solvent.

The dilution factor for 1 ml of coffee into 9 ml of water is:

$\frac{\text{Amount transferred}}{\text{Total amount}}=\frac{\text{Amount transferred}}{\text{(amount transferred}+\text{amount water added)}}$

Now put all the given values in this expression, we get the dilution factor for coffee.

$\frac{\text{Amount transferred}}{\text{(amount transferred}+\text{amount water added)}}=\frac{1}{(1+9)}=\frac{1}{10}=0.01$

Therefore, the coffee diluted was, 1 in 10, or 1/10, or 0.1 dilution factor.

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He following balanced equation shows the formation of water. 2H2 + O2 mc020-1.jpg 2H2O How many moles of oxygen (O2) are require

mrs_skeptik [129]

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A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

number of moles of $NH_4Cl$ = 0.40 mol/L × 200 × 10⁻³L

= 0.08 mole

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= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

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Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

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$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

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8 0

2 years ago

2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0

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