All categories

3 years ago

Question 3

Chemistry

1 answer:

Dmitrij [34]3 years ago

4 0

Answer:

snow and very low temperatures are predicted for tomorrow.

Explanation:

snow and very low temperatures are associated with average weather condition of a place thus making it a statement about climate.

You might be interested in

2. What is the mass of 16.3 L of helium gas?

lilavasa [31]

(16.3 L) / (22.414 L/mol) x (4.0026 g He/mol) = 2.91 g

3 0

3 years ago

Read 2 more answers

The gas pressure inside a container decreases when which of the following happens? the number of molecules is increased and the

ASHA 777 [7]

<h2>can't understand your question..... can u explain me May be I can help u!</h2><h2 /><h2 />

7 0

3 years ago

Is weathering long term or short term and why

Marrrta [24]

Long term because if you leave something out to be weathered then it can’t be unweathered because of the drastic change of the object.

4 0

3 years ago

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

Xelga [282]

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :

The relation between the rate constant the activation energy is,

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

8 0

3 years ago

Read 2 more answers

A substance is analyzed and found to contain 85.7% carbon and 14.3% hydrogen by weight. A gaseous sample of the substance is fou

atroni [7]

Answer:

The empirical formula of the compound is CH2

Explanation:

Step 1: Data given

A substance contains 85.7 % carbon and 14.3 % hydrogen.

The substance has a density of 1.87 g/L

1 mol occupies 22.4 L

Molar mass of carbon = 12 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate molar mass of the substance

Since 1 mol occupies 22.4 L;

1 mol of this substance = 1.87g/L *22.4 = 41.888 grams

This means the molar mass of the substance is 41.888 g/mol

Step 3: Calculate mass of carbon:

85.8 % is carbon

this means 41.888 * 0.858 = 35.94 grams

Step 4: Calculate moles of carbon

moles C = mass C/ Molar mass C

Moles C = 35.94 grams / 12 g/mol

Moles C = 2.995 moles

Step 5: Calculate mass of hydrogen:

14.3 % is hydrogen

this means 41.888 * 0.143 = 5.99 grams

Step 6 :Calculate moles of hydrogen

Moles H = 5.99 grams / 1.01 g/mol

Moles H = 5.93 moles

Step 7: Calculate mol ratio

Ratio C:H = 1:2

The empirical formule = CH2

Step 8: calculate molar formule

Molar mass of empirical formule = 14.02 g/mol

n = Molar mass of substance / molar mass of empirical formule

n = 41.888 / 14.02 = 3

This means we have to multiply the empirical formula by 3

3*(CH2) = C3H6

C3H6 can be propene or cyclopropane

5 0

3 years ago