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3 years ago

COPY out the following sentences and fill in the gaps.

Chemistry

1 answer:

malfutka [58]3 years ago

4 0

Answer:

1. Hydrogen ions; acidic

2. Alkali; hydroxide ions; alkaline

3a. Sulfuric acid --> 2 Hydrogen ions + sulfate ion

H₂SO₄ --> 2H+ + SO₄²-

3b. Sodium hydroxide --> Sodium ion + Hydroxide ion

NaOH --> Na+ + OH-

Explanation:

1. Sulfuric acid releases hydrogen ions in solution. This makes the solution acidic.

Acids produce hydrogen ions when dissolved in aqueous solutions.

2. Sodium hydroxide is an alkali. It releases hydroxide ions in solution. This makes the solution alkaline.

Alkalis are soluble bases that produce hydroxide ions in solution.

3a. Sulfuric acid --> 2 Hydrogen ions + sulfate ions

H₂SO₄ --> 2H+ + SO₄²-

The equation above is for the ionization of sulfuric acid

b. Sodium hydroxide --> Sodium ion + Hydroxide ion

NaOH --> Na+ + OH-

The equation above is for the ionization of sodium hydroxide

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KIM [24]

Answer:

Li + ZnCO3

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3 years ago

Read 2 more answers

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 $NO_{2}_{(g)}$

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