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antiseptic1488 [7]
3 years ago
15

OMG PLEASE HELP!

Chemistry
1 answer:
Paraphin [41]3 years ago
4 0
Assuming the temperature is 20° C,
100 g of water dissolves 34 g,
200 g of water will dissolve: 200 x 34 / 100
= 68 g

The third option is correct.


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The most common shape for a galaxy is ______.
andrew-mc [135]
The least common shape for a galaxy is cluster .
8 0
3 years ago
I searched it up but it keeps saying chemical energy, but in the option doesn’t have it...I’m torn between B and C...helppppp!!!
jeyben [28]

Answer: The correct answer is in chemical bonds

Explanation:

When coal is burnt, these components burn and release energy. The energy released is by the chemical reaction between the constituents and oxygen

8 0
3 years ago
Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
Leya [2.2K]

Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

3 0
3 years ago
A sample that contains only SrCO3 and BaCO3 weighs 0.800 g. When it is dissolved in excess acid, 0.211 g car- bon dioxide is lib
MrRissso [65]

Answer:

53.9%

Explanation:

1 mole of BaCO₃ yields  1 mole of CO₂,

1 mole of SrCO₃ yields    1 mole of CO₂

m₁ = mass of BaCO₃

m₂ =  mass SrCO₃

molar mass of SrCO₃  = 147.63 g/mol

molar mass of  BaCO₃ = 197.34 g/mol

molar mass of CO₂ = 44.01 g/mol

mole of CO₂ in 0.211 g = 0.211 g / 44.01 = 0.00479

mole of BaCO₃ = m₁ / 197.34

mole of SrCO₃  = m₂ / 147.63

mole of BaCO₃ + mole of SrCO₃  = 0.00479

(m₁ / 197.34) + (m₂ / 147.63) = 0.00479

147.63 m₁ + 197.34 m₂ = 139.55

m₁ + m₂ = 0.8

m₁ = 0.8 - m₂

147.63 (0.8 - m₂) + 197.34 m₂ = 139.55

118.104 - 147.63 m₂ + 197.34 m₂ = 139.55

49.71 m₂ = 139.55 - 118.104 = 21.446

m₂ = 21.446 / 49.71 = 0.431

the percentage of m₂ ( SrCO₃ ) in the sample = 0.431 / 0.8 = 0.539 × 100 = 53.9%

5 0
4 years ago
An aqueous HCL solution has a proton concentration equal to 6.00 mol/L. The HCL concentration in this solution is ————M.
Harman [31]

Answer:

HCl conc.= 6.0mol/L

Explanation:

From the dissociation of HCl= 1 mole H+ and 1mol Cl-, which is equivalent stoichiometrically in concentration to that of 1 mol HCl,

4 0
4 years ago
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