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ipn [44]
3 years ago
5

Keri has to identify a mysterious brown liquid in science class. She pours it through a filter, but the substance looks the same

. Then she heats the substance and collects the water that evaporates. The water droplets look the same as the original substance. Keri concludes that since the substance cannot be separated it must be a compound. Is Keri correct?
Chemistry
2 answers:
Rasek [7]3 years ago
6 0

Answer is: No, because the substance could be an element.

Pure substance is made of only one type of atom (element) or only one type of molecule, it has definite and constant composition with distinct chemical properties.

Pure substances can be separated chemically, not physically, that is difference between pure substances and mixtures.

Elements (for example copper, iron, sulfur) and compounds (water, sodium chloride) have definite and constant composition with distinct chemical properties.

Allisa [31]3 years ago
4 0

The answer is NO. It can not be a compound it is an element.

In a component mixture it can be separated physically way from the other.

In element it can be only separated by a way of destroying the compound itself by chemical reaction.

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When 0.100 mol of carbon is burned in a closed vessel with8.00
antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of O_2 = 8.00 g

Molar mass of O_2 = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of O_2.

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2C+O_2\rightarrow 2CO

From the balanced reaction we conclude that

As, 2 mole of C react with 1 mole of O_2

So, 0.1 moles of C react with \frac{0.1}{2}=0.05 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and C is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO

From the reaction, we conclude that

As, 2 mole of C react to give 2 mole of CO

So, 0.1 moles of C react to give 0.1 moles of CO

Now we have to calculate the mass of CO

\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO

\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g

Therefore, the mass of carbon monoxide form can be 2.8 grams.

5 0
3 years ago
a 20.5g sample of cleaning detergent contains 8.61g of NH40H.CALCULATE the percentage composition of nitrogen in the cleaning de
AysviL [449]

Answer:

The mass percentage composition of nitrogen in the sample of the cleaning detergent is approximately 16.78%  

Explanation:

The given mass of the sample of the cleaning detergent, m₁ = 20.5 g

The mass of the ammonium hydroxide, NH₄OH in the detergent, m₂ = 8.61 g

The molar mass of NH₄OH = 35.04 g/mol

The molar mass of nitrogen, N = 14.01 g/mol

Therefore, the mass, m₃ of nitrogen, N, in 8.61 g of ammonium hydroxide, NH₄OH, is found as follows;

m₃ = (14.01/35.04) × 8.61 g = (402,087/118,800) g ≈ 3.44 g

The mass of nitrogen, N, in the ammonium hydroxide, NH₄OH, contained in the 20.5 g sample of the cleaning agent, m₃ ≈ 3.44 grams

The percentage composition of nitrogen in the sample of the cleaning detergent, %N is given as follows;

\% Composition = \dfrac{Mass \ of \ component}{Total \ mass \ of \ cleaning \ detergent} \times 100

Therefore;

%N ≈ ((3.44 g)/(20.5 g)) × 100 ≈ 16.78 %

The percentage composition of nitrogen, %N ≈ 16.78%.

6 0
3 years ago
What element is Column 1, period 3
Finger [1]

Answer:

Sodium (Na)

Explanation:

The element on the periodic table at Column (group) 1, period 3 is Sodium (Na)

3 0
3 years ago
Calorimeter containing 1000 grans of water the initial temperature of water is 24.85 degrees . the heat of water is 4.184 j/g de
kvasek [131]
<h3><u>Answer;</u></h3>

10.80 ° C

<h3><u>Explanation;</u></h3>

From the information given;

Initial temperature of water =  24.85°C

Final temperature of water = 35.65°C

Mass of water = 1000 g

The specific heat of water ,c = 4.184 J/g °C.

The heat capacity of the calorimeter  = 695 J/ °C

Change in temperature ΔT = 35.65°C - 24.85°C

                                             = 10.80°C

3 0
3 years ago
How many grams of nh3 can be produced from 3.72 mol of n2 and excess h2.
olganol [36]
N₂       + 3H₂ ⇒ 2NH₃
1mol       :         2mol
3,72mol  :         7,44mol

n = 7,44mol
M = 17g/mol

m = n * M = 7,44mol * 17g/mol = 126,48g
7 0
3 years ago
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