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maksim [4K]
3 years ago
13

Which is NOT part of a normal microscope? a.ocular b. nosepiece c.stage d.objective

Chemistry
2 answers:
kozerog [31]3 years ago
6 0

Answer:

Nose piece

Explanation:

I have never seen one on a normal microscope

Evgesh-ka [11]3 years ago
3 0

Answer:

I believe it is a Nose piece

Explanation:

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ludmilkaskok [199]

Answer:Poop

Explanation:poop

7 0
3 years ago
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
3 years ago
Sex-linked disorders occur more frequently in males than females because males have two X chromosomes and females have two Y chr
Vilka [71]
This is false because males have 1 X and 1 Y chromosome. It's females who have 2 X chromosomes. So, it's false.
5 0
3 years ago
Read 2 more answers
Identify three nonmetal atoms that, when bonded covalently, have an argon electron configuration.
Advocard [28]

Answer:

Phosphorus

Sulphur

And chlorine are non metals when bonded covalently gain electronic configuration of argon

Like PCl3

SO2 and HCl

Explanation:

8 0
2 years ago
The molarity (M) of an aqueous solution containing 29.8 g of sucrose, C12H22O11, in 120 mL of solution is:
Rama09 [41]

Answer:

0.726 mol·L⁻¹

Step-by-step explanation:

c = moles/litres

=====

Moles = 29.8 × 1/342.30

Moles = 0.087 06 mol

=====

Litres = 120 × 1/1000

Litres = 0.120 L

=====

c = 0.087 06/0.120

c = 0.725 mol·L⁻¹


5 0
3 years ago
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