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horrorfan [7]
3 years ago
14

What is the momentum of a 0.01 kg toy car moving at 5.0 m/s?

Chemistry
1 answer:
snow_tiger [21]3 years ago
8 0

Answer: 0.05 kgm/s

Explanation:

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(6.21×10^3)(0.1050) <br><br>(Scientific Notation)
mel-nik [20]
6.21 x 10^3 = (Move decimal point 3 spaces to the right)

6210

6210 (0.1050)

652.05
3 0
3 years ago
A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.
MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

A.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

B.

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

C.

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0
3 years ago
The process in which the nucleus of an atom splits into two lighter atoms, releasing a large amount of energy, is nuclear .
IgorC [24]

It's nuclear fission

6 0
3 years ago
Read 2 more answers
A sample of gas has a volume of 215 cm3 at 23.5 degrees Celsius and 84.6kPa what volume will the gas occupy at stp
miss Akunina [59]

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

P1=84.6 kpa

T1=23.5 +273=296.5 K

V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

5 0
3 years ago
Read 2 more answers
The immediate electron acceptor for the majority of the oxidative reactions of the citric acid cycle is.
Anna [14]

NAD serves as the bulk of the oxidative processes in the citric acid cycle's initial electron acceptor.

<h3>What are electron acceptors in citric acid cycle?</h3>
  • In the Krebs cycle, which transfers electrons via the electron transport chain with oxygen as the final acceptor, coenzymes like FAD and NAD+ are reduced.
  • In a single cycle, three NADH+ and one FADH2 are produced, and when the cycle enters the electron transport chain, 10 ATP is produced.
  • The final electron acceptor in the electron transport chain is oxygen. The proton gradient in the intermembrane gap is produced by NADH molecules donating electrons that are then transmitted through a number of different proteins.
<h3>What occurs throughout the citric acid cycle?</h3>

The cycle of citric acid: In the citric acid cycle, a six-carbon citrate molecule is created when an acetyl group from acetyl CoA is joined to a four-carbon oxaloacetate molecule.

Citrate is oxidized over a number of steps, generating two molecules of carbon dioxide for each acetyl group added to the cycle.

learn more about  citric acid cycle here

<u>brainly.com/question/14900762</u>

#SPJ4

6 0
2 years ago
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