Answer plzzzz im begging

The chemical mixture that composes our atmosphere is called ________.

Virty [35]

Answer: Air

Hope this helps

3 0

3 years ago

Carbon diselenide (CSe2) is a liquid at room temperature. The normal boiling point is 125°C, and the melting point is –45.5°C. C

Tema [17]

Answer:

Explanation: The strengths of the inter molecular forces varies as follows -

$CO_{2}< CS_{2} < CSe_{2}$

The normal boiling point of CSe2 is 125°C and that of CS2 is 116°C, which explains the trend that as we move down the group, the boiling point of e compound increases as the size increases.

This usually happens because larger and heavier atoms have a tendency to exhibit greater inter molecular strengths due to the increase in size . As the size increases, the valence shell electrons move far away from the nucleus, thus has a greater tendency to attract the temporary dipoles.

And larger the inter molecular forces, more tightly the electrons will be held to each other and thus more thermal energy would be required to break the bonds between them.

5 0

3 years ago

Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe

jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

3 0

2 years ago

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Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13

Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

$\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\$

$=\frac{1571}{10}\\\\=157.1$

Calculating the standardn:

$\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\$

Please find the correct equation in the attached file.

$=28.195$

For point a:

$=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\$

For point b:

$=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M$

For point c:

$= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\ M$

It is calculated by using the slope value that is $1.75 \times 10^9 M^{-1}$ . The slope value $1.75 \times 10^9 M^{-1}$ is ambiguous.

7 0

2 years ago

How many moles of sodium cyanide (NaCN) would be needed to produce 4.2 moles of sodium sulfate (Na2SO4)? H2SO4 + 2NaCN → 2HCN +

leonid [27]

The purpose of a chemical equation is to relate the amounts of reactants to the amounts of products based on the rate each is consumed. In this problem, one mole of sulfuric acid is consumed along with two moles of sodium cyanide to produce two moles of hydrocyanic acid and one mole of sodium sulfate. The relationship between sodium cyanide and sodium sulfate is 2:1, meaning that two moles of NaCN is required to produce one mole of sodium sulfate.

To produce 4.2 moles of sodium sulfate, two times this amount of NaCN is required. This means that you would need 8.4 moles of sodium cyanide.

Hope this helps!

4 0

3 years ago

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