CHEMISTRY HIGH SCHOOL

11x + 13
12x+6 Solve for x

Answers

Answer 1

Answer:

Answer:

x = 7

Explanation:

Simplifying

11x + 13 = 12x + 6

Reorder the terms:

13 + 11x = 12x + 6

Reorder the terms:

13 + 11x = 6 + 12x

Solving

13 + 11x = 6 + 12x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-12x' to each side of the equation.

13 + 11x + -12x = 6 + 12x + -12x

Combine like terms: 11x + -12x = -1x

13 + -1x = 6 + 12x + -12x

Combine like terms: 12x + -12x = 0

13 + -1x = 6 + 0

13 + -1x = 6

Add '-13' to each side of the equation.

13 + -13 + -1x = 6 + -13

Combine like terms: 13 + -13 = 0

0 + -1x = 6 + -13

-1x = 6 + -13

Combine like terms: 6 + -13 = -7

-1x = -7

Divide each side by '-1'.

x = 7

Simplifying

x = 7

Answer 2

Answer:

Answer:Simple and best practice solution for 11x+13=12x+6 equation. ... + 6 Reorder the terms: 13 + 11x = 6 + 12x Solving 13 + 11x = 6 + 12x Solving for variable 'x'.

Explanation:

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Which statement explains why water is classifiedas a compound? (1) Water can be broken down by chemical means. (2) Water is a liquid at room temperature. (3) Water has a heat of fusion of 334 J/g. (4) Water is a poor conductor of electricity.

Use electron transfer or electron shift to identify what is oxidized and what is reduced in each reaction :a) 2Na(s) + Br2(l) ----> 2NaBr(s)
b) H2(g) + Cl2(g) ----> 2HCl(g)
c) 2Li(s) + F2(g) ----> 2LiF(s)
d) S(s) + Cl2(g) ----> SCl2(g)
e)N2(g) + 2O2(g) ----> 2NO2(g)
f) Mg(s) +Cu(NO3)2(aq) = Mg(NO3)2(aq) + Cu(s)

For each reaction above, identify the reducing agent and the oxidizing agent

Answers

Answer :

Oxidation-reduction reaction : It is a type of reaction in which oxidation and reduction reaction occur simultaneously.

Oxidation reaction : It is the reaction in which a substance looses its electrons. In the oxidation reaction, the oxidation state of an element increases.

Reduction reaction : It is the reaction in which a substance gains electrons. In the reduction reaction, the oxidation state of an element decreases.

(a) The balanced chemical reactions is,

$2Na(s)+Br_2(l)\rightarrow 2NaBr(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Na\rightarrow Na^(1+)+1e^-$

Reduction : $Br_2+2e^-\rightarrow 2Br^(1-)$

From this we conclude that, 'Na' is oxidized and $'Br_2'$ is reduced in this reaction. The reducing agent is, 'Na' and oxidizing agent is, $'Br_2'$ .

(b) The balanced chemical reactions is,

$H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

Half reactions of oxidation and reduction are :

Oxidation : $H_2\rightarrow H^(1+)+1e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^(1-)$

From this we conclude that, $'H_2'$ is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, $'H_2'$ and oxidizing agent is, $'Cl_2'$ .

(c) The balanced chemical reactions is,

$2Li(s)+F_2(g)\rightarrow 2LiF(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Li\rightarrow Li^(1+)+1e^-$

Reduction : $F_2+2e^-\rightarrow 2F^(1-)$

From this we conclude that, 'Li' is oxidized and $'F_2'$ is reduced in this reaction. The reducing agent is, 'Li' and oxidizing agent is, $'F_2'$ .

(d) The balanced chemical reactions is,

$S(s)+Cl_2(g)\rightarrow SCl_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $S\rightarrow S^(2+)+2e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^(1-)$

From this we conclude that, 'S' is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, 'S' and oxidizing agent is, $'Cl_2'$ .

(e) The balanced chemical reactions is,

$N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $N_2\rightarrow N^(4+)+4e^-$

Reduction : $O_2+4e^-\rightarrow 2O^(2-)$

From this we conclude that, $'N_2'$ is oxidized and $'O_2'$ is reduced in this reaction. The reducing agent is, $'N_2'$ and oxidizing agent is, $'O_2'$ .

(f) The balanced chemical reactions is,

$Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Mg\rightarrow Mg^(2+)+2e^-$

Reduction : $Cu^(2+)+2e^-\rightarrow Cu$

From this we conclude that, $'Mg'$ is oxidized and $'Cu'$ is reduced in this reaction. The reducing agent is, 'Mg' and oxidizing agent is, 'Cu'.

a) Na is oxidised Br is reduced
b) H is oxidised Cl is reduced
c) Li is oxidised F is reduced
d)S is oxidised Cl is reduced
e) N is oxidised O is reduced
f) Mg is oxidised and N is reduced

Remember: Oxidation= loss and Reduction= gains

If 37.4 grams of water decompose at 297 Kelvin and 1.30 atmospheres, how many liters of oxygen gas can be produced? Show all of the work used to solve this problem. 2H2O (l) yields 2H2 (g) + O2 (g)

Answers

To determine the liters of oxygen gas, we convert the 37.4 grams of water to moles, which is 2. 078 moles. The moles oxygen is half the moles of water which is equal to 1.039 moles. Using ideal gas equation, PV=nRT, we get the volume from the given temperature and pressure. The volume is 19.49 liters.

The pressure of 1 mol of gas is decreased to 0.5 atm at 273 K. What happens to the molar volume of the gas under these conditions?

Answers

Answer : The molar volume of the gas will be, 44.82 L

Solution :

Using ideal gas equation,

$PV=nRT$

where,

n = number of moles of gas = 1 mole

P = pressure of the gas = 0.5 atm

T = temperature of the gas = 273 K

R = gas constant = 0.0821 Latm/moleK

V = volume of the gas.

Now put all the given values in the above equation, we get the molar volume of the gas.

$(0.5atm)* V=(1mole)* (0.0821Latm/moleK)* (273K)$

$V=44.82L$

Therefore, the molar volume of the gas will be, 44.82 L

Consider the mixture of ethanol, c2h5oh, and o2. how many molecules of co2, h2o, c2h5oh, and o2 will be present if the reaction goes to completion?

Answers

The reaction between ethanol and oxygen must be a combustion reaction to give carbon dioxide and water as products.

The reaction between ethanol, $C_2H_5OH$ and oxygen, $O_2$ is as follows:

$C_2H_5OH+O_2\rightarrow CO_2+H_2O$

In order to balance the reaction (the number of atoms of each element in the reaction are same on both the sides that is product and reactant), $O_2$ is multiplied by 3 in the reactant side and $CO_2$ is multiplied by 2 and $H_2O$ is multiplied by 3 on product side.

So, the balanced reaction is:

$C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O$

From the balanced reaction it is clear that 1 mole of $C_2H_5OH$ reacts with 3 moles of $O_2$ to give 2 moles of $CO_2$ and 3 moles of $H_2O$ .

Since, 1 mole = $6.022* 10^(23)$ molecules (Avogadro's number)

So,

Number of molecules of $C_2H_5OH$ = $6.022* 10^(23)$ molecules.

Number of molecules of $O_2$ = $6.022* 10^(23)* 3 = 18.066* 10^(23)$ molecules.

Number of molecules of $CO_2$ = $6.022* 10^(23)* 2 = 12.044* 10^(23)$ molecules.

Number of molecules of $H_2O$ = $6.022* 10^(23)* 3 = 18.066* 10^(23)$ molecules.

Answer : The number of molecules of $CO_2,H_2O,C_2H_5OH\text{ and }O_2$ are 2, 3, 1 and 3 respectively.

Explanation :

The given balanced chemical reaction is,

$C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O$

Balanced chemical reaction is the reaction in which the number of atoms of an individual elements present on reactant side must be equal to the product side.

The species present on the left side of the right arrow is the reactant and the species present on the right side of the right arrow is the product.

From the balanced chemical reaction, we conclude that there are 1 molecule of $C_2H_5OH$ , 3 molecules of $O_2$ , 2 molecules of $CO_2$ and 3 molecules of $H_2O$ .

6) At elevated temperatures, methylisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN):CH3NC (g)  CH3CN (g)At the start of an experiment, there are 0.200 mol of reactant and 0 mol of product in the reaction vessel.After 25 min, 0.108 mol of reactant (CH3NC) remain. There are mol of product (CH3CN) in thereaction vessel.A) 0.200 B) 0.540 C) 0.022 D) 0.092 E) 0.308

Answers

Answer: The moles of product present in the vessel is 0.092 moles

Explanation:

For the given chemical equation:

$CH_3CN(g)\rightleftharpoons CH_3CN(g)$

We are given:

Initial moles of reactant = 0.200 moles

Unreacted moles of reactant = 0.108 moles

Reacted moles of reactant = 0.200 - 0.108 = 0.092 moles

By Stoichiometry of the reaction:

1 mole of reactant produces 1 mole of product

So, 0.092 moles of reactant will produce = $\frac{1}[1}* 0.092=0.092mol$ of product

Hence, the moles of product present in the vessel is 0.092 moles

What is ionization energy? a. the minimum energy required to remove one electron from an atom.
b. the maximum energy required to remove one electron from an atom.
c. the maximum energy required to move one electron to the valence shell.
d. the minimum energy required to remove one electron from another electron.

Answers

Answer:

D) energy required to remove a valence electron

Explanation:

Ionization energy can be defined as the atom's ability to and the energy required to remove a valence electron. Ionization energy increases up a group and to the right along a period. So helium has the highest first Ionization energy meaning that it is very hard to remove an electron from helium because octet is already complete.

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11x + 1312x+6 Solve for x

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11x + 13
12x+6 Solve for x