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3 years ago

PLEASE HELP ME!

Chemistry

1 answer:

kifflom [539]3 years ago

4 0

Explanation:

a. Magnesium oxide is used mainly in soil treatment and groundwater remediation, wastewater treatment, etc. it is used for it's acid buffering capacity and effectiveness in dissolving heavy metals.

b. MgO is the chemical equation for the reaction.

c. multiply the given mass of O2 by the inverse of it's molar mass. multiply the molar ratio (from the balanced equation) between O2 and MgO. Multiply by the molar mass of MgO. 32g O2 x 1 mol O2 ... 32g O2 x 2mol MgO 1mol O2 x 40g MgO. .... 1mol MgO = 80 g.

d. Mg (s) + 2 HCl (aq) produces MgCl 2(aq) + H 2 (g). Where the letter "s" stands for solids, and "g" is for gas and "aq" represents aqueous solution.

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24 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!

erastovalidia [21]

I'd say b, but i'm not 100 percent sure.

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3 years ago

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How can you tell if an element has been oxidized or reduced in a reaction?

fomenos

Gaining of electrons is reduction and loss of electrons is oxidation.

Hope this helped! :)

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4 years ago

Which substance is a binary acid

Bumek [7]

It is known that acids compounds contains hydrogen and produces hydrogen ion in water. A binary acid however is an acid that have two elements, one of the element has a hydrogen attached to it. Examples of binary acids are hydrogen fluoride (HF), hydrogen bromide (HBr) and hydrogen sulfide (H2S). In naming a binary acid, it has two rules; one, as pure compounds and two, as acid solutions. For pure compounds, start with the name ‘hydrogen’ and end the anion name with ‘-ide’. For acidic compounds, start with ‘hydro-‘, end the anion with ‘-ic’ and add ‘acid’.

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3 years ago

Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.

ruslelena [56]

Answer:

a) 0.168 moles O2

b) 9.50 grams O2

c) 0.01662 kg NO

d)88.9 %

Explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen = 6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = 0.168 moles O2

b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol = 9.50 grams O2

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = 0.01662 kg NO

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

Calculate moles of NH3 = 2.51 grams / 17.03 g/mol = 0.147 moles

Calculate moles of O2 = 3.76 grams / 32 g/mol = 0.118 moles

Determine the limiting reactant

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

Calculate moles H2O: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

Calculate mass H2O = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

Calculate % yield = (2.27/2.552)*100 % = 88.9 %

4 0

3 years ago

You are going to standardize your sodium hydroxide by titrating with potassium hydrogen phthalate. As an example, you dissolve 0

Kay [80]

Answer:

0.13 M

Explanation:

The reaction equation is;

NaOH(aq) + KHC8H4O4(aq) ------> KNaC8H4O4(aq) + H2O(l)

Molar mass of KHP = 204.22 g/mol

Amount of KHP= mass/ molar mass = 0.3365 g/204.22 g/mol = 1.65 × 10^-3 moles

n= CV

Where;

C= concentration

V= volume in dm^3

n= number of moles

C= n/V = 1.65 × 10^-3 moles × 1000/250 = 6.6 × 10^-3 M

If 1 mole of KHP reacts with 1 mole of NaOH

1.65 × 10^-3 moles of KHP will react with 1.65 × 10^-3 moles of NaOH

From

n= CV

We have that only 12.44 ml of NaOH reacted

C= n/V = 1.65 × 10^-3 moles × 1000/12.44

C= 0.13 M

At the equivalence point, the KHP solution turned light pink.

4 0

3 years ago