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3 years ago

46g of a mineral contained 16g copper, 14g iron and 16g sulphur. Calculate the empirical

Chemistry

1 answer:

pochemuha3 years ago

7 0

Answer:

CuFeS2

Explanation:

Calculate the moles of each substance by doing moles= mass/relative atomic mass. you should get 0.25 moles of copper and iron and 0.5 moles of sulfur. Then divide all of those numbers by 0.25 (as its the lowest value) you should get 1 for copper and iron and 2 for sulfur. This represents the ratio that they are in within the mineral.

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If liquid water ___ energy it will become ice

kati45 [8]

Answer:

the answer is 1- loses

Explanation:

When water freezes it gives up some of the water's energy.

6 0

3 years ago

CsBr formula name???

olganol [36]

Answer

PubChem CID/molecular formula

Explanation:

Cesium bromide

PubChem CID 24592

Molecular Formula CsBr or BrCs

Synonyms CESIUM BROMIDE 7787-69-1 Caesium bromide Cesiumbromide Cesium bromide (CsBr) More...

Molecular Weight 212.81 g/mol

Component Compounds CID 260 (Hydrogen bromide) CID 5354618 (Cesium)

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4 0

3 years ago

A buffer is prepared by adding 300. 0 ml of 2. 0 mnaoh to 500. 0 ml of 2. 0 mch3cooh. what is the ph of this buffer? ka= 1. 8 10

Anton [14]

The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.

<h3>What is the Henderson-Hasselbalch equation?</h3>

Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.

Given,

Acid disassociation constant (ka) = 1. 8 10⁻⁵

Concentration of NaOH = 2.0 M

Concentration of CH₃COOH = 2.0 M

pKa value is calculated as,

pKa = -log Ka

pKa = - log (1. 8 x 10⁻⁵)

Substituting the value of pKa in the Henderson-Hasselbalch equation as

pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]

pH = - log (1. 8 x 10⁻⁵) + log [1]

= 4.745 + 0

= 4.75

Therefore, 4.75 is the pH of the buffer.

Learn more about the Henderson-Hasselbalch equation here:

brainly.com/question/27751586

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6 0

1 year ago

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)

lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

$\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C$