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Eduardwww [97]
3 years ago
12

What is the term used to describe a change in motion?

Chemistry
1 answer:
Irina-Kira [14]3 years ago
8 0

Answer:

it should be momentum if im not mistaken

You might be interested in
Fill in the blanks with appropriate options given in below:
jarptica [38.1K]

Answer:

1) -COOH

2) -NH2

3) hydrogen bonds

4) dispersion forces

5) -CH3

6) hydrogen bonds

7) negative

8) negative

9) positive

Explanation:

Alanine has a <u>-COOH</u> and a <u>-NH2</u> group available to form <u>hydrogen bonds</u> with water molecules.

Although there are some potential <u>dispersion forces</u> between the terminal <u>-CH3</u> group of alanine and hexane molecules, we expect the <u>hydrogen bonds</u> between alanine and water to be stronger.

Stronger intermolecular attractive forces between alanine and water lead to a more <u>negative ΔHmix</u> and more <u>negative (smaller positive)</u> ΔHsoln for water than for hexane.

4 0
3 years ago
Elements in what family are all gases?
Alina [70]

Answer:

Noble gases.

Good Luck & Have a good day!

(.❛ ᴗ ❛.)

6 0
3 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
How are covalent compounds similar to acids?
Ivan

Answer:

Acids are substances that produce an create high amounts of H+ ions when dissolved in water. And because the hydrogen bond's with non-metals,it forms covalent bonds. So,all acids are covalent bonds.

Explanation:

Colavent compounds are colavent bonds

Source:

https://www.quora.com/Are-acids-covalent-compound

5 0
3 years ago
Caffeine, a stimulant found in coffee and soda, hasthe mass percent composition: C. 49.48%, H, 5.19%. N. 28.85% 0. 16.48% The mo
borishaifa [10]

We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

C_4H_5O_1N_2

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

C_8H_{10}O_2N_4

8 0
1 year ago
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