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3 years ago

Help! I need help on how to do these problems.

Chemistry

1 answer:

svetlana [45]3 years ago

6 0

Answer:

a. 3; b. 5; c. 10; d. 12

Explanation:

pH is defined as the negative log of the hydronium concentration:

pH = -log[H₃O⁺] (hydronium concentration)

For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.

For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:

pOH = -log[OH⁻] (hydroxide concentration)

Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.

Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.

(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)

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Describe the method of steel manufacture used when removes the impurities carbon, phosphorus and sulfur

melisa1 [442]

Explanation:

Steelmaking is the process of producing steel from iron ore and/or scrap. In steelmaking, impurities such as nitrogen, silicon, phosphorus, sulfur and excess carbon (most important impurity) are removed from the sourced iron, and alloying elements such as manganese, nickel, chromium, carbon and vanadium are added to produce different grades of steel.

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6 0

3 years ago

How many formula units are there in 450 g of Na2S04?

Mashutka [201]

Formula units in 450 g of $Na_{2} So_{4}$ is 1.93 × 10²⁴ formula units.

Explanation:

First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

$$\frac{450 g}{142.04 g/mol} = 3.2 moles$

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.

3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.

So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.

4 0

3 years ago

What is the relationship between the following compounds??

34kurt

And what are the compounds?

8 0

4 years ago

Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338

Elden [556K]

Answer: The concentration of chloride ions in the solution obtained is 0.674 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KCl:

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

$0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol$

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

For magnesium chloride:

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

$0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol$

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = $(2\times 0.224)=0.448mol$

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

$\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M$

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

5 0

3 years ago

Consider four elements from Group 17: fluorine in the second period, chlorine in the third period, bromine in the fourth period,

nika2105 [10]

Answer:

A. Fluorine

Explanation:

I got it right on Plato.

6 0

3 years ago