Ask question

Ask question

All categories

2 years ago

12

Amol has divided some substances into two groups as shown in the table below. He wants to further separate all the substances in

Group 1 again into exactly two groups. Which of the following criteria will allow him to do so?
A. liquids and solids

B. mixtures and pure substances

C. naturally occurring and man-made substances.

D. substances completely soluble and insoluble in water.

2 answers:

Nonamiya [84]2 years ago

6 0

The criteria that will allow Amol separate group 1 substances into two groups are mixtures and pure substances (option B).

<h3>What are mixtures and pure substances?</h3>

Mixtures are substances that are made up of two or more substances that can be easily separated.

Pure substances, on the other hand, are elements and compounds that cannot be separated by chemical means.

In group 1 of the above table, the substances can be classified as follows:

Soil - mixture
Tomato ketchup - mixture
Smoke - compound
Milk - mixture
Hair shampoo - mixture

Therefore, the criteria that will allow Amol separate group 1 substances into two groups are mixtures and pure substances.

Learn more about mixture at: brainly.com/question/24898889

#SPJ1

Degger [83]2 years ago

5 0

Answer:

Explanation:

b is the right answer

You might be interested in

what happens to the yield of hydrogen if the reaction is carried out at a higher temperature but at the same pressure

valkas [14]

Answer: It increases the rate of desulfurization at a constant feed rate

Explanation:

5 0

4 years ago

(a) the characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustio

ella [17]

By stoichiometry and assume that:

CxH2xOy + zO2 -> xCO2 + xH2O

<span>
CO2: 9.48/44 = 0.215 mmol
H2O: 3.87/18 = 0.215 mmol
mass of C = 0.215 * 12 = 2.58 mg
mass of H = 0.215 * 2 * 1 = 0.43 mg
mass of O in ethylbutyrate = 4.17 - 2.58 - 0.43 = 1.11 mg
So C/O = 2.58/1.11 ≈ 3 </span>

<span>
Thus we have C3H6O</span>

<span> </span>

8 0

3 years ago

Read 2 more answers

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

Taya2010 [7]

Answer:

- Empirical:

$C_3H_7$

- Molecular:

$C_6H_{14}$

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

$n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC$

Moreover, the moles of hydrogen:

$n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH$

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

$C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}$

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

$C_3H_7$

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

$C_6H_{14}$

Which is hexane.

Best regards.

6 0

4 years ago

A tank of gas is found to exert 8.6 atm at 38°C. What would be the required

Vesna [10]

Answer:

36.2 K

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 8.6 atm

Initial temperature of the gas (T₁): 38°C

Final pressure of the gas (P₂): 1.0 atm (standard pressure)

Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

6 0

3 years ago

Please help fast<br> PLease

vaieri [72.5K]

i believe the answer is a Brainiest plz

6 0

4 years ago