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3 years ago

How many Cl- ions are there in 400.5 grams of AlCl3?

Chemistry

2 answers:

den301095 [7]3 years ago

6 0

Answer:

The answer to your question is letter A

Explanation:

Process

1.- Calculate the molecular mass of AlCl₃

Al = 1 x 27 = 27

Cl = 3 x 35.5 = 106.5

AlCl₃ = 27 + 106.5 = 133.5

2.- Find the number of ions of Cl using proportions

133.5 grams of AlCl₃ ------------- 3 ions of Chlorine

400.5 grams of AlCl₃ ------------ x

x = (400.5 x 3) / 133.5

x = 9

ankoles [38]3 years ago

4 0

Answer:

Explanation:

the answer is d because when you solve it's d

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(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

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

$V_e = 2178 \ L$

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allochka39001 [22]

ΔH2 = - δH1 δH2 = - 2 x δH1 δH2 = 2 x <span>δ</span>H1

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SVETLANKA909090 [29]

Answer:

B) They will react because X and Y can share two pairs of electrons to become stable

Explanation:

The electron configurations of two elements x and y are given :

X: 1s2 2s2 2p6

Y: 1s2 2s2 2p6 3s2 3p6

The statement that is true for both the elements is that, they both will react as they both can share two pairs of electrons to become stable.

To become stable the outermost shell or p orbital should have 8 electrons, so element X can gain 2 atoms to become stable.

Element Y can also react as it can also share two atoms to fulfill its 3p orbital and will stable.

Hence, the correct option is "B".

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Ronch [10]

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From the balanced reaction we conclude that

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3 years ago