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2 years ago

A runner can run at a speed of 4km/h in 2 hours. What distance will she

Chemistry

1 answer:

damaskus [11]2 years ago

3 0

Answer:

Distance cover = 8 km

Explanation:

Given:

Speed of runner = 4 km/h

Time taken = 2 hour

Find:

Distance cover

Computation:

Distance = Speed x Time

Distance cover = Speed of runner x Time taken

Distance cover = 4 x 2

Distance cover = 8 km

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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

2 years ago

Why would you add boiling chips/stones to a solution that is to be refluxed? when should you add them?

anygoal [31]

Answer: Boiling chips provide surfaces on which bubbles can form as the liquid boils.

Hope this helps!

6 0

2 years ago

Which statement is true about the electrons that can be located together in an orbital?

Salsk061 [2.6K]

The two electrons that share an orbital repel each other.

All electrons bear a negative charge. They are held in their orbits by the attractive force of charged protons. The farther away an orbital is to the atomic nucleus the easier it is to expunge an electron from this distant orbital shell.

Explanation:

Because electrons have the same negative charge, they repel each other especially when they occupy the same orbital shell in an atom. To reduce this repulsion, each of the electrons in the orbital shell (remember electrons occupy orbital shells of atoms in 2s) assumes an opposite quantum (Ms) spin; one with – ½ while the other + ½ .

Learn More:

For more about electrons check out;

brainly.com/question/13251728

brainly.com/question/13174681

#LearnWithBrainly

3 0

2 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

2 years ago

What is the purpose of the lab? Weather patterns

Sedaia [141]

Answer:

Here:

Explanation:

To familiarize students with experimental apparatus, the scientific method, and methods of data analysis so that they will have some idea of the inductive process by which the ideas were originated. To teach how to make careful experimental observations and how to think about and draw conclusions from such data.

7 0

2 years ago