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3 years ago

I need helppppppppppp

Chemistry

2 answers:

agasfer [191]3 years ago

8 0

Answer:

option c hope it will help u

kicyunya [14]3 years ago

7 0

A B C

a. Mg(12) 2.8.2 -2e- = Mg2+

b. S(16) 2.8.6 +2e- = S2-

c. MgS

d. No, because it is neutral

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When the [H+] in a solution is 1.7 × 10−9 M, what is the pOH

Harrizon [31]

The pOH of a solution is related to the concentration of hydroxide ions in a solution. Also, it is related to the pH of the solution since the sum of pOH and pH is equal to 14. We can use this relation to solve this problem. We do as follows:

pH = -log[<span>1.7 × 10−9 M</span>] = 8.8
pOH + pH = 14
pOH + 8.8 = 14
pOH = 5.2

7 0

4 years ago

Read 2 more answers

1. Which of the following compounds is most likely to exist as a covalent molecule? O NaBr O KNO, O Bel, C O CH_NH2?

DIA [1.3K]

Answer:

nabeo

Explanation:

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3 years ago

What is a salt? A compound formed from a(n) ____ from an acid and a(n) ____ from a base.

zloy xaker [14]

Answer:

In chemistry, a salt is an ionic compound that can be formed by the neutralization reaction of an acid and a base. Salts are composed of related numbers of cations and anions so that the product is electrically neutral.

Explanation:

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3 years ago

Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO an

Ivenika [448]

Answer:

The remain gases are $o_{2}_{(g)}$ and $NO_{2}_{(g)}$

Pressure of $O_{2}_{(g)}$ $1.09 atm O_{2}_{(g)}$

Pressure of $NO_{2}_{(g)}$ $1.09 atm NO_{2}_{(g)}$

Explanation:

We have the following reaction

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}$

Now we calculate the limit reagents, to know which of the two gases is completely depleted and which one is in excess.

Excess gas will remain in the tank when the reagent limits have run out and the reaction ends.

To calculate the limit reagent, we must calculate the mols of each substance. We use the ideal gas equation

$PV= nRT$

We cleared the mols

$n=\frac{PV}{RT}$

PV=nrT

replace the data for each gas

Constant of ideal gases

$R= 0.082\frac{atm.l}{mol.K}$

Transform degrees celsius to kelvin

$25+273=298K$

$NO_{g}$

$n=\frac{0.500atm.3.90l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.080mol NO_{(g)} \\$

$O_{2}_{(g)$

$n=\frac{1atm.2.09l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.086mol O_{2}_{(g)} \\$

Find the limit reagent by stoichiometry

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}$

$0.086mol O_{2}_(g).\frac{2mol NO_{(g)} }{1mol O_{2}_{(g)} } =0.17mol NO_{(g)}$

Using $O_{2}_{(g)}$ as the limit reagent produces more $NO_{(g)}$ than I have, so oxygen is my excess reagent and will remain when the reaction is over.

$NO_{(g)}$

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{1mol O_{2}_ {(g)} }{2mol NO_{(g)} } =0.04mol O_{2}_{(g)}$

Using $NO_{(g)}$ as the limit reagent produces less $O_{2}_{(g)}$ than I have, so $NO_{(g)}$ is my excess reagent and will remain when the reaction is over.

Calculate the moles that are formed of $NO_{2}_{(g)}$

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{2mol NO_{2}_ {(g)} }{2mol NO_{(g)} } =0.080mol NO_{2}_{(g)}$