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HACTEHA [7]
3 years ago
14

What mass of Na2SO4 is needed to make 2.5L of 2.0 M solution?

Chemistry
1 answer:
Gala2k [10]3 years ago
5 0
Moles of Na2SO4 = conc. x volume(dm3)
= 2 x (2500cm3/1000)
= 5 mol


Mass of Na2SO4 = moles x molar mass
= 5 x ((23 x 2) + 32.1 + (16 x 4))
= 710.5g
You might be interested in
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
HELP MEE please with this question this question is 100pts and i will give branliest i only have 5 min left guys thanx:) Which s
LenaWriter [7]

Answer:

The cell grows into its full size

The cell copies it’s dna

Explanation:

hope this helps pls mark brainliest

8 0
2 years ago
Which ones are right to this question?
Vitek1552 [10]

The average Kenectic energy

7 0
3 years ago
Here is the electron configuration for Magnesium. How many total electrons are in the 2nd energy level?
balandron [24]

Answer:

8 electrons

Explanation:

Magnesium is present on group 2.

It has 2 valence electrons.

Electronic configuration of magnesium:

Mg₁₂ = 1s² 2s² 2p⁶ 3s²

1st energy level contain 2 electrons.(1s²)

2nd energy  level contain 8 electrons. (2s² 2p⁶)

3rd energy level contain 2 electrons. (3s²)

3rs energy level of magnesium is called valence shell. It contain two valance electrons. Magnesium can easily donate its two valance electrons and get stable electronic configuration.

It react with halogens and form salt. For example,

Mg + Cl₂   →   MgCl₂

7 0
3 years ago
I need help ASAP! How many reactants and products are in each element according to the chemical equation above.
Len [333]

Answer:

O 4

C 1

1 sodium

3 H

Products

1 sodium

3 H

4 O

1 C

Explanation:

7 0
3 years ago
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