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Mila [183]
2 years ago
11

How many total atoms are there is a sugar molecule

Chemistry
1 answer:
Tju [1.3M]2 years ago
7 0

Answer:

The white stuff we know as sugar is sucrose, a molecule composed of 12 atoms of carbon, 22 atoms of hydrogen, and 11 atoms of oxygen (C12H22O11). Like all compounds made from these three elements, sugar is a carbohydrate

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1.00 M CaCl2 Density = 1.07 g/mL
Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

m = 1 mol\times 111 g/mol = 111 g

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0
3 years ago
Do Alkali Metal Compounds form precipitates? And Why?
Alexeev081 [22]

Answer:

Alkali metal hydroxides can be used to test the identity of metals in certain salts. The colour of the precipitate will help identify the metal : Calcium hydroxide is soluble; no precipitate is formed.

6 0
2 years ago
Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in
Veronika [31]

Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is  2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\   \frac{m_{S} }{m_{O} }= 2

Let us express it as

SO_{\frac{m_{S} }{m_{O} }} = 2

For SO₂,

Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

\frac{m_{S} }{m_{O} }= \frac{32}{(2)(16)}\\\\   \frac{m_{S} }{m_{O} }= 1

Let us express it as

SO_2_{\frac{m_{S} }{m_{O} }}= 1

Now, for the ratio of both the above-calculated ratios,

\frac{SO_{\frac{m_{S} }{m_{O} }}}{SO_2_{\frac{m_{S} }{m_{O} }}}=\frac{2}{1}

The required ratio is 2:1

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2 years ago
Wave y and wave z are two types of electromagnetic waves traveling through a vacuum suppose the frequency of wave y is about thr
Feliz [49]
Frequency and energy have a direct relationshipWave Y transfers about three times the amount of energy as wave Z because energy and frequency have a direct relationship
4 0
3 years ago
Read 2 more answers
Chemical equations consist of the chemical formulas of the reactants on the right
Dahasolnce [82]
False. Chemical products are on the right side.
6 0
2 years ago
Read 2 more answers
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