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3 years ago

80.0 grams of NaOH is dissolved in water to make a 6.00 L solution what is the molarity of the solution

Chemistry

1 answer:

Elanso [62]3 years ago

3 0

Answer:

0.33 mol/ L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer will help you. Have a nice day !

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Organic acids are weaker than inorganic once it is true or false

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false

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A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t

pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = $pK_a=9.31$

First we have to calculate the moles of HCN and NaCN.

$\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole$

and,

$\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole$

The balanced chemical reaction is:

$HCN+NaOH\rightarrow NaCN+H_2O$

Initial moles 0.1116 0.0461 0.08978

At eqm. (0.1116-0.0461) 0 (0.08978+0.0461)

0.0655 0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$pH=9.31+\log (\frac{0.1359}{0.0655})$

$pH=9.63$

Therefore, the pH of the solution is, 9.63

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Layers of rock containing fossils, like the layers illustrated here, are MOST LIKELY composed of ______________ rocks. A) igneou

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3 years ago

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

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3 years ago