CHEMISTRY HIGH SCHOOL

Calculate the number of atoms of each type in C12H22O11 and record

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Answer 1

Answer:

Answer:

1 mol of sucarose has 7.22 x10²⁴ atoms of C, 1.32 x10²⁵ atoms of H and 6.62 x10²⁴ atoms of O

Explanation:

C12H22O11 is the molecular formular for sucarose

So in 1 mol of sacarose, we have 12 mols of carbon, 22 moles of hydrogen, and 11 mols of oxygen.

As you know, 1 mol has 6.02x10²³ atoms so these are the rule of three to calculate the number of atoms of each type.

1 mol ______ 6.02x10²³

12 moles ______ 12 . 6.02x10²³ = 7.22 x10²⁴

22 moles ______ 22 . 6.02x10²³ = 1.32 x10²⁵

11 moles _______ 11 . 6.02x10²³ = 6.62 x10²⁴

Answer:

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METALLIC BOND:

Metallic Bond is basically a electrostatic force of attraction between the metal ions which are arranged in a lattice(Lattice is a regular repeating pattern) and the free electrons floating around the metal ions.

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PROPERTIES OF METALLIC BOND:

1- Metals are good conductors of heat and electricity

2- Metals are ductile

3- Metals are malleable

4- Metals are solid at room temperature

5- Metals Possess Metallic luster

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A solution contains 20.0 g of C6H12O6 (glucose) in 250 g. of water.A. what is freezing point depression of solvent?

Answers

Answer:

The freezing point depression of solvent is 0.8265 K.

Explanation:

Mass of glucose = 250 g

Mass of solvent = 20.0 g = 0.020 kg

Molal depression water constant of water, $K_f$ = 1.86 K kg/mol

Molality of the solution = $(20 g)/(180 g/mol* 0.250 kg)=0.4444 mol/kg$

Normal freezing point of water = $T=273 K$

Freezing point of solution = $T_f$

$\Delta T_f=T-T_f$

$\Delta T_f=K_f* molality=1.86 K kg/mol * 0.4444 mol/kg=0.8265 K$

The freezing point depression of solvent is 0.8265 K.

Molar mass glucose = 180 g/mol

number of moles:

20 / 180 = 0.111 moles

Δ t f * ( Kf) *( i ) * ( m )

250 g of water in Kg = 250 / 1000 = 0.25 kg

1.86 ºC/m * 0.111 / 0.25

1.86ºC/m * 0.444

= 0.82584ºC

Fp = 0ºC - 0.82584

Fp = - 0.82584ºC

hope this helps!

Which structural adaptation would help a plant survive better in a shady environment? A.
thorns

B.
small leaves

C.
large leaves

D.
brightly colored flowers

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Answer:C.

large leaves

Explanation: I took the test

The function of most proteins depends primarily on the

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type and order of amino acids

1.type and order of amino acids. 2.environment of the organism.3.availability of starch molecules.

The percent composition by mass of certain compound is 35.5 percent carbon, 4.8 percent hydrogen, 37.9 percent oxygen, 8.3 percent nitrogen, and 13.5 percent sodium. What is the molecular formula if the molar mass is roughly 170 grams per mole

Answers

Answer:

$C_(5)H_(8)NNaO_(4)$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 35.5

Molar mass of C = 12.0107 g/mol

% moles of C = $(35.5)/(12.0107)$ = 2.95569

% of H = 4.8

Molar mass of H = 1.00784 g/mol

% moles of H = $(4.8)/(1.00784)$ = 4.76266

% of N = 8.3

Molar mass of N = 14.0067 g/mol

% moles of N = $(8.3)/(14.0067)$ = 0.59257

% of Na = 13.5

Molar mass of Na = 22.989769 g/mol

% moles of N = $(13.5)/(22.989769)$ = 0.58721

% of O = 37.9

Molar mass of O = 15.999 g/mol

% moles of O = $(37.9)/(15.999)$ = 2.36889

Taking the simplest ratio for C, H, O, N, Na and O as:

2.95569 : 4.76266 : 0.59257 : 0.58721 : 2.36889

= 5 : 8 : 1 : 1 : 4

The empirical formula is = $C_(5)H_(8)NNaO_(4)$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12*5 + 1*8 + 14 + 23 + 16*4 = 169 g/mol

Molar mass = 170 g/mol

So,

Molecular mass = n × Empirical mass

170 = n × 169

⇒ n ≅ 1

The molecular formula = $C_(5)H_(8)NNaO_(4)$

Answer : The molecular formula of a compound is, $C_5H_8O_4NNa$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 35.5 g

Mass of H = 4.8 g

Mass of O = 37.9 g

Mass of N = 8.3 g

Mass of Na = 13.5 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Molar mass of N = 14 g/mole

Molar mass of Na = 23 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (35.5g)/(12g/mole)=2.96moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (4.8g)/(1g/mole)=4.8moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (37.9g)/(16g/mole)=2.37moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (8.3g)/(14g/mole)=0.593moles$

Moles of Na = $\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= (13.5g)/(23g/mole)=0.587moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $(2.96)/(0.587)=5.0\approx 5$

For H = $(4.8)/(0.587)=8.1\approx 8$

For O = $(2.37)/(0.587)=4.0\approx 4$

For N = $(0.593)/(0.587)=1.0\approx 1$

For Na = $(0.587)/(0.587)=1$

The ratio of C : H : O : N : Na = 5 : 8 : 4 : 1 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_5H_8O_4N_1Na_1=C_5H_8O_4NNa$

The empirical formula weight = 5(12) + 8(1) + 4(16) + 1(14) + 1(23) = 169 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=(170)/(169)=1$

Molecular formula = $(C_5H_8O_4NNa)_n=(C_5H_8O_4NNa)_1=C_5H_8O_4NNa$

Therefore, the molecular of the compound is, $C_5H_8O_4NNa$

Determine the number of neutrons in an atom of Si-29

Answers

15 as 29 (atomic mass-number of neutrons and protons) - 14 (atomic number-number of protons)

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Calculate the number of atoms of each type in C12H22O11 and record

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Answer:

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METALLIC BOND:

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PROPERTIES OF METALLIC BOND:

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