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3 years ago

CH3(CH2)50CH3 How many atoms

Chemistry

1 answer:

ahrayia [7]3 years ago

3 0

Explanation:

An atom is the smallest indivisible particle of any matter. Atoms are distinct and distinguishable from one another and are called elements:

CH₃(CH₂)₅0CH₃

There are 3 different kinds of atoms in the compound above called

elements

Carbon

Hydrogen

Oxygen

In the compound the subscripts gives the number of atoms:

C - 7

H - 16

0 - 1

Number of atoms in the compound is 24.

Learn more:

Number of atoms brainly.com/question/10419836

#learnwithBrainly

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A hydrogen filled balloon has a volume of 8.3 L at 36°C and 751 mmHg. How many moles of hydrogen are inside the balloon?

tamaranim1 [39]

Answer: I think the formula is PV=nRT and I divide both sides by RT, but this is as far as I can get in my equation before I get stumped: (751 mm Hg) (8.3 L)/ (309 K) Can you help?

Explanation:

8 0

2 years ago

2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s

Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.

In order to prepare 10 mL, 10 μM; 4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM; 6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM; 8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

= 2 mL

2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

6 0

2 years ago

If 200. g of water at 20°C absorbs 41 840 J of energy, what will its final temperature be? (Specific Heat of water is 4.184 J/g*

Elena-2011 [213]

Answer: The final temperature will be $70^0C$

Explanation:

To calculate the specific heat of substance during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed =41840 J

c = specific heat = $4.184J/g^0C$

m = mass of water = 200 g

$T_{final}$ = final temperature =?

$T_{initial}$ = initial temperature = $20^0C$

Now put all the given values in the above formula, we get:

$41840J=200g\times 4.184J/g^0C\times (T_{final}-20)^0C$

$T_{final}=70^0C$

Thus the final temperature will be $70^0C$

3 0

2 years ago

For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

6 0

3 years ago

From a 2.875 g sample containing only iron, sand, and salt, 0.660 g of iron and 1.161 g of sand were separated and recovered. Wh

Serjik [45]

Answer:

36.66%

Explanation:

Step 1: Given data

Mass of iron: 0.660 g

Mass of sand: 1.161 g

Mass of the sample: 2.875 g

Mass of salt: ?

Step 2: Calculate the mass of salt

The mass of the sample is equal to the sum of the masses of the components.

m(sample) = m(iron) + m(sand) + m(salt)

m(salt) = m(sample) - m(iron) - m(sand)

m(salt) = 2.875 g - 0.660 g - 1.161 g

m(salt) = 1.054 g

Step 3: Calculate the percent of salt in the sample

We will use the following expression.

%(salt) = m(salt) / m(sample) × 100%

%(salt) = 1.054 g / 2.875 g × 100% = 36.66%

7 0

3 years ago