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Olenka [21]
2 years ago
12

how does the percentage by mass of the solute describe the concentration of an aqueous solution os potassium sulfate

Chemistry
1 answer:
kirza4 [7]2 years ago
3 0

Answer:

Explanation:

In an aqueous solution of potassium sulfate (K₂SO₄), the solute is K₂SO₄ and the solvent is water. The percentage by mass describes the grams of solute there are dissolved per 100 grams of solution. It can be calculated as:

mass percentage = (mass of solute/total mass of solution) x 100%

For example, in an aqueous solution which is 2% by mass of K₂SO₄, there are 2 grams of K₂SO₄ per 100 g of solution.

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The empirical formula of glucose is CH2O the molecular formula of glucose is 6 times than the empirical formula. what is the mol
jeka57 [31]

Answer:

The molecular formula of glucose is C₆H₁₂O₆

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of smallest whole number of atoms.

Molecular formula:

It gives the total number of atoms in a molecule of compound.

The molecular formula and empirical formula can be related as follow:

Molecular formula = n × empirical formula

Given data:

Empirical formula = CH₂O

Molecular formula = ?

It is stated in given problem that molecular formula is the 6 times of the empirical formula.

Molecular formula = n × empirical formula

Molecular formula = 6 × CH₂O

Molecular formula = C₆H₁₂O₆

The molecular formula of glucose is C₆H₁₂O₆.

5 0
3 years ago
Which is the electron configuration for zinc?
alisha [4.7K]

[AR] 3d10 4s2 is the configuration for zinc

3 0
3 years ago
Read 2 more answers
Sodium hydroxide, NaOHNaOH; sodium phosphate, Na3PO4Na3PO4; and sodium nitrate, NaNO3NaNO3, are all common chemicals used in cle
nadya68 [22]

Answer:

NaOH > Na3PO4 > NaNO3

Explanation:

in NaOH

23/40 * 100 = 57.5%

in Na3PO4

3 * 23/164 * 100 = 42%

In NaNO3

23/85 * 100 = 27.1%

Hence;

NaOH > Na3PO4 > NaNO3

4 0
3 years ago
If you have 6 moles of reactant A and excess of B and C, how much product E would be formed
pychu [463]
B and C are in excess so amount of E will be determined by A. 

Amount of product is determined by limiting reagents - Always.

Hence 6 moles of E will be formed.

Hope this helps!
3 0
3 years ago
How many molecules are there in 4.00 l of oxygen gas at 500.∘ c and 50.0 torr?
Oliga [24]
From  the  ideal  gas law   
pv=nRT , n  is  therefore PV/RT
R is  the
R is  gas  constant =62.364 torr/mol/k
P=500torr
 V=4.00l
T=500+273=773k
n={(500 torr x 4.00l)/(62.364 x773k)}=0.041moles
the  number  of  molecules=moles  x avorgadro costant that is  6.022x10^23)
6.022 x 10^23)  x0.041=2.469 x10^22molecules
3 0
3 years ago
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