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3 years ago

What would be the pH of an H2SO4 solution if the [H+] = 4.58 x 10-5 moles/liter?

Chemistry

1 answer:

Ghella [55]3 years ago

4 0

Answer: pH of an $H_2SO_4$ solution is 4.34

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$H_2SO_4\rightarrow 2H^++SO_4^{2-}$

Putting in the values:

$pH=-\log[4.58\times 10^{-5}]$

$pH=4.34$

Thus pH of an $H_2SO_4$ solution if the $[H^+]$ is $4.58\times 10^{-5}]$ is 4.34

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How many moles are in 55 g or NaCk

seraphim [82]

Answer:

Explanation:

Ok, so I suppose you meant NaCl. But ok:

So we have the formula moles = grams/MM, where MM is molar mass. The molar mass of NaCl is 58.44 g/mol. That means that 55/58.44 is the number of moles which appears to be 0.94 moles.

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3 years ago

The energy to move water And allow it to change from a to a gas or solid ultimately comes from the

rosijanka [135]

Answer:

Explanation:

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3 years ago

A scientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. How many grams of na3po4

natima [27]

Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 $\frac{mol}{L}$

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL

$Volume of solution = 725 mL * \frac{1L }{1000 mL}$

Volume of solution = 0.725 L

Following steps can be done to find mass of :

Step 1 : Write the dissociation reaction of Na₃PO₄ .

$Na_3PO_4 3 Na^+ + PO_4^3^-$

Step 2: Find moles of Na⁺ ions :

Mole of Na⁺ ions can be calculated using molarity formula which is : $Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }$

Plugging value of Molarity and volume

$1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}$

Multiplying both side by 0.725 L

$1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L$

Mole of Na⁺ ions = 0.725 mol

Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :

Mole ratio is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion = 3

Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3

Step 4 : To find mole of Na₃PO₄

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

$Mole of Na_3PO_4= Mole of Na^+ * Mole ratio$

$Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }$

Mole of Na₃PO₄ = 0.242 mol

Step 5 : To find mass of Na₃PO₄

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :

$Mass (g) = mole (mol) * molar mass \frac{g}{mol}$

$Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}$

Mass of Na₃PO₄ = 39.619 g

Step 6 : To round off mass of Na₃PO₄ to correct sig fig .

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g

8 0

3 years ago

The irreversible elementary gas-phase reaction is carried out isothermally at 305 K in a packed-bed reactor with 100 kg of catal

tamaranim1 [39]

Answer:

0.856.

Explanation:

Lets represent the irreversible elementary gas phase equation of reaction as

A + B -----------------------------------> C + D

We have that the percentage of conversion is 80%.

The pressure, p from the ratio of exit pressure and entering pressure is p = 2/20 = 1/10 = 0.1.

Therefore, n = 1 - p^2/ weight of the catalyst = 1 - 0.1^2/ 100 = 9.9 × 10^-3 kg cat^-.

Now, let's make use of the equation below;

J/ 1 - J = kb^2/ u [ w - nw^2/2] ----------(1).

0.8 / 1- 0.8 = k ( 0.4)^2/ 10 [ 100 - (9.9× 10^-3 × 100^2/ 2] .

k = 4.95 dm^6/ kg.cat .mol.min

The turbulent flow= 1/2 × 9.9 × 10^-3 = 4.95 × 10^-3 kg cat^-.

Thus, making use of the equation (1) again, we have that;

{4.95 × 10^-3 × 0.4}/ 10 × [ 100 - (4.95 × 10^-3 × 100^2)] / 2 = 5.964.

Therefore, a/1 - a = 5.964.

5.964( 1 - a) = a.

5.964 - 5.964a = a.

5.964 = a + 5.964a.

5.964 = 6.964a.

a = 5.964/ 6.964 = 0.856.

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3 years ago

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nydimaria [60]

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Explanation:

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3 years ago