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3 years ago

Which describes the composition of matter in the figure C?

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

3 0

Answer:

The answer to your question is Mixture

Explanation:

Element is an substance whose atoms are all the same because all the atoms

have the same number of protons, neutrons and electrons.

Compound is a substance formed when 2 or more atoms combined.

Mixture is a substance formed by 2 or more different kinds of atomos or

compounds.

Phase is a state of matter like solid, liquid or gas.

According to the definitions, we conclude that the figure C is a Mixture.

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Density of 30.00 mL of H2O

ohaa [14]

its not possible sorry

6 0

3 years ago

SOMEONE PLEASE HELP

Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

$= 8.16 \times 10^{ - 6} \: kg$

$\boxed{density = \frac{mass}{volume} }$

Density of iron bar

$= \frac{0.0642}{8.16 \times 10^{ - 6} }$

= 7870 kg/m³ (3 s.f.)

$\boxed{mass = density \: \times volume}$

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

$= 12 \times 10^{ - 6} \: m^{3} \\ = 1.2 \times 10^{ - 5} \: m ^{3}$

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

$= \frac{0.151} {1.25 \times 10^{ - 5} } \\ = 12600 \: kg/ {m}^{3}$

(3 s.f.)

6 0

3 years ago

9. How many grams are in 2.0 x 105 molecules of carbon monoxide?

timama [110]

Answer: 9.3 x 10^ 18 g CO

Explanation:

Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:

(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO

This cancels molecules CO.

Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:

(1 mol CO) x 28 g CO / 1 mol CO

This cancels mol CO, which leaves grams CO.

5 0

3 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life

mixas84 [53]

T is amount after time t
Ao is initial amount 
t is time 
HL is half life 

log (At) = log [ Ao x (1/2)^(t/HL) ] 
log (At) = log Ao + log (1/2)^(t/HL) 
log (At) = log Ao + (t/HL) x log (1/2) 

( log At - log Ao) / log (1/2) = t / HL 
log (At/Ao) / log (1/2) = t / HL 

HL = t / [( log (At / Ao)) / log (1/2) ] 

HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] 

HL = 14.4 s / 2 = 7.2 seconds

8 0

3 years ago