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suter [353]
3 years ago
14

Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneo

us:_______. (a) rice pudding, (b) seawater, (c) magnesium, (d) crushed ice.
Chemistry
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance

Explanation:

Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.

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7 0
3 years ago
Identify two acids and two bases that you use or come into contact with in an average week. Identify uses for each substance.
skelet666 [1.2K]

Explanation:

Two acids we come into contact with in an average week

  1. Vinegar is an 10% solution of acetic acid (CH_3COOH)in water. Used in salad dressing and while cooking food. It has a sour taste.
  2. Citric acid present in fruits and vegetables like : lemons, orange, tomatoes etc. It is a weak organic acid with sour taste.

Two bases we come into contact with in an average week.  

  1. Baking soda (NaHCO_3) is used in baking food like: cakes, cookies, breads. Baking soda is one of the ingredient while baking breads and cakes.
  2. Caustic soda (NaOH) is used for preparation of detergents, papers , soaps etc. We use soaps and detergents for washing.
4 0
2 years ago
Read 2 more answers
How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles
sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

.

8 0
11 months ago
How many grams of water do you need to weigh if a reaction requires 5 moles of h2o
rewona [7]

One mole of water weighs 18 grams. H₂O is composed of 2H= 2 and 1 0=16 adding gives you 18. number of moles= mass/ Relative Molecular Mass

Therefore, mass= Relative Molecular Mass×number of moles

                           = 18×5 moles

                           = 90 grams

5 0
3 years ago
f 23.6 mL of 0.200 M NaOH is required to neutralize 10.00 mL of a H3PO4 solution , what is the concentration of the phosphoric a
Eduardwww [97]

Answer:

Explanation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

mole of NaOH = 23.6 * 10 ⁻³L * 0.2M

= 0.00472mole

let x be the no of mole of H3PO4 required of  0.00472mole of NaOH

3 mole of NaOH required ------- 1 mole of H3PO4

0.00472mole of NaOH ----------x

cross multiply

3x = 0.0472

x = 0.00157mole

[H3PO4] = mole of H3PO4 / Vol. of H3PO4

= 0.00157mole / (10*10⁻³l)

= 0.157M

<h3>The concentration of unknown phosphoric acid is  0.157M</h3>
7 0
2 years ago
Read 2 more answers
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