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3 years ago

Explain why mass cannot be used as a property to identify a sample of matter.

1 answer:

4 0

Mass cannot be used as a property to identidy a sample of matter because it is an extensive property which depends on the amount of matter in a sample, not the tupe of matter it is.

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1718 l of a 0.3556-m c3h7oh solution is diluted to a concentration of 0.1222 m, what is the volume of the resulting solution

Agata [3.3K]

Answer : The volume of the resulting solution will be 0.4999 L.

Explanation : As the two molar concentration of the $C_{3} H_{7}OH$ solution are given and one of the volume concentration needs to be found.

So, according to the formula :- $m_{1}V_{1} = m_{2}V_{2}$

And here if we consider $V_{1}$ as 0.1718 L moles and $m_{1}$ as 0.3556 moles then we need to find $V_{2}$ as $m_{2}$ is given as 0.1222 moles.
Hence on solving the formula we get,

$V_{2}$ = (0.1718 X 0.3556) / 0.1222 = 0.4999 L

5 0

3 years ago

This is the chemical formula for carvone (the chemical that gives spearmint its flavor): An organic chemist has determined by me

xz_007 [3.2K]

The given question is incomplete. The complete question is :

This is the chemical formula for carvone (the chemical that gives spearmint its flavor) $C_{10}H_{14}O$ : An organic chemist has determined by measurements that there are 27.3 moles of carbon in a sample of carvone. How many moles of oxygen are in the sample

Answer: 2.73 moles of oxygen are there in the sample.

Explanation:

The chemical formula of carvone is $C_{10}H_{14}O$ .

This means:

When 10 moles of carbon are there in carvone , 1 mole of oxygen is there in carvone

Thus for 27.3 moles of carbon in carvone , $\frac{1}{10}\times 27.3=2.73$ mole of oxygen is there in carvone.

Thus 2.73 moles of oxygen are there in the sample.

6 0

2 years ago

What is the correct balanced equation for the following equilibrium constant expression?

weeeeeb [17]

The correct answer is 3) 2CO2(g) ⇄ 2CO(g) + O2(g)

this is the correct one because it is a decomposition reaction and all the number of atoms is equal on both sides.

there are 2 C atoms on both sides.

and 4 O atoms on both sides.

and 1) the atoms numbers are equal on both sides but not correct as it not a

correct number as it has 1/2 O2.

and 2) CO2(g) ⇆ CO(g) + O2

the number of O atoms is not equal on both sides of the equation.

we have 2 O atoms on the left side and 3 O atoms on the right side.

so, this not a balanced equation.

4) also not correct 2CO(g) + O2 ⇆ 2CO2

as it is not a decomposition reaction and the 2CO & O2 are as reactants not products.

so the correct answer is 3) 2CO2(g) ⇆ 2CO(g) + O2(g)

8 0

3 years ago

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago

15. How many moles of carbon tetrachloride (CCI) is represented by 543.2 g of carbon tetrachloride? The atomic weight of carbon

sladkih [1.3K]

Answer:

well, first off. the formula for carbon tetrachloride is CCl4

We need to find the molar mass of carbon tetrachloride

1(Mass of C) + 4(mass of chlorine)

1(12) + 4(35.5)

12 + 142

154 g/mol

Number of moles of CCl3 in 543.2g CCl3

n = given mass / molar mass

n = 543.2/153

n = 3.53 moles

always remember to brainly the questions you find helpful

4 0

3 years ago