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3 years ago

Help please!

1 answer:

4 0

Answer: In factories they burn coal under a giant container(a really big gumbo pot pretty much). when the fire is hot enough it starts to boil. When it boils it creates steam and the steam blows on a fan that makes electricity.

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If the temperature of a gas increases, describe what should happen to the pressure of the gas. Assume the volume and amount of g

Sliva [168]

The pressure will increase with decreasing volume. if they remain constant, that is.

4 0

3 years ago

Which of the following represents the mass of 1 molecule of SH2?

mr_godi [17]

Answer:
5.645 × 10⁻²³ g

Solution:

Step 1) Calculate Molar Mass of SH₂;

Atomic Mass of Sulfur = 32 g/mol

Atomic Mass of H₂ = 2 g/mol
--------------------
Molecular Mass of SH₂ = 34 g/mol

Step 2: Calculate mass of one molecule of SH₂ as;

As,

Moles = # of Molecules / 6.022 × 10²³

Also, Moles = Mass / M.Mass So,

Mass/M.mass = # of Molecules / 6.022 × 10²³

Solving for Mass,

Mass = # of Molecules × M.mass / 6.022 × 10²³

Putting values,

Mass = (1 Molecule × 34 g.mol⁻¹) ÷ 6.022 × 10²³

Mass = 5.645 × 10⁻²³ g

5 0

3 years ago

Read 2 more answers

7. Propane, C₂H₁, has become a popular gas to use in fireplaces.

kramer

Answer:

.......................

8 0

2 years ago

Which of the following statements is true regarding both theories and laws?

Mariana [72]

Answer:

D. Supported by observations

Explanation:

Because theories change over time, they both need to be proven, they do not go off of consensus, but they do need to be supported by observations.

HOPE IT HELPS❣️

6 0

3 years ago

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=> smallest whole number ratio of elements in a compound

Molecular Formula => actual 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

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

=> 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