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3 years ago

How many grams of k2so4 would you need to prepare 1500 g of 5.0% k2so4 solution?

1 answer:

5 0

Data Given:
% w/w = 5 %

Solution weight = 1500 g

Solute weight = ?

Formula Used:
% w/w = (Mass of Solute / Mass of Solution) × 100

Solving for Mass of Solute,

Mass of Solute = (% w/w × Mass of Solution) ÷ 100

Mass of Solute = (5 × 1500 g) ÷ 100

Mass of Solute = 75 g K₂SO₄

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Which element is a liquid at 1000 k 1.Ag 2.Al 3.Ca 4.Ni

AlexFokin [52]

in these options not any Element are liquid at 1000k i think you type wrong in first option It is Hg not Ag and Hg is liquid at 1000 k

7 0

3 years ago

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Given:

bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ $= \frac{59.7}{16} = 3.73125 \ moles$

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0

3 years ago

The _____________ is the temperature at which a substance changes from solid to liquid; ________________ is the temperature at w

Aleksandr-060686 [28]

The _____melting point________ is the temperature at which a substance changes from solid to liquid; _______boiling point_________ is the temperature at which a substance changes from a liquid to as gas; _______vapourisation_________ is the process by which atoms of molecules leave a liquid and become a gas.

5 0

3 years ago

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha

NARA [144]

There are 2.32 x 10^6 kg sulfuric acid in the rainfall.

Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M

By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4

We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4

3 0

4 years ago

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What’s the distance from the nucleus of an atom to its theoretical outer edge of electron orbits

ollegr [7]

Answer:

The range of atoms = (30-300 pm) depending upon the element

Explanation:

The Atomic radii of the atom is the distance from the center of the circle to the outermost orbital.

The center of the circle is the nucleus and the radii is the outermost boundary.

The actual size of the atom is decided on the basis of the Zeff . Also known as effective nuclear charge.

Zeff: It is the net positive charge felt by the outermost electron by the nucleus.

The value of Zeff depends upon the shielding constant. More the shielding less will be the Zeff . Hence the size of the atom increases.

Due to shielding the outermost electrons feel less pull of nucleus.

The greater the Zeff , the smaller the radius of the atom.

The formula used to calculate the atomic mass is :

$r_{n}=\frac{52.9n^{2}}{Z}$ pm

Here "pm"= picometers

$1 pm = 10^{-12}m$

The size of the smallest atom H-atom = 120 pm

The range of atoms = (30-300 pm)

4 0

3 years ago