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3 years ago

Which set of numbers gives the correct possible values of I for n = 3?

Chemistry

1 answer:

Mnenie [13.5K]3 years ago

7 0

Answer:

0 1 2 3

Explanation:

L=0 1 2 3....(n-1)

determine the shape of atomic orbitals

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How many pili are in a bacterial cell

Snowcat [4.5K]

Answer: Bacterial species where observed Typical number on cell Distribution on cell surface

Escherichia coli (common pili or Type 1 fimbriae) 100-200 uniform

Neisseria gonorrhoeae 100-200 uniform

Streptococcus pyogenes (fimbriae plus the M-protein) ? uniform

Pseudomonas aeruginosa 10-20 polar

Explanation:

Pili are structures that extend from the surface of some bacterial cells.

These are hollow, non-helical, filamentous appendages.

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7 0

3 years ago

Can someone please help me ?

DochEvi [55]

Answer:

The acceleration of ball is 18 m/s².

Explanation:

Definition:

The acceleration is rate of change of velocity of an object with respect to time.

Formula:

a = Δv/Δt

a = acceleration

Δv = change in velocity

Δt = change in time

Units:

The unit of acceleration is m.s⁻².

Acceleration can also be determine through following formula,

F = m × a

a = F/m

Given data:

Mass of ball = 0.60 kg

Force on it = 10.8 N

Acceleration = ?

Solution:

a = 10.8 N/0.60 kg

a = 18 N/kg

N = kg.m/s²

a = 18 kg.m/s² / kg

a = 18 m/s²

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4 years ago

What is the diet of sloth bears ?

Svetllana [295]

This is what I got hope it helps

4 0

3 years ago

Read 2 more answers

Calculate the number of grams of solute needed to make each of the following solutions:

AleksAgata [21]

Explanation:

(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

1) 100 g of 0.500% (w/w) NaI

Mass of solution = 100 g

Mass of solute = x

Required w/w % of solution = 0.500%

$0.500\%=\frac{x}{100 g}\times 100$

$x=\frac{0.500\times 100 g}{100}=0.500 g$

0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.

2) 250 g of 0.500% (w/w) NaBr

Mass of solution = 250 g

Mass of solute = x

Required w/w % of solution = 0.500%

$0.500\%=\frac{x}{250 g}\times 100$

$x=\frac{0.500\times 250 g}{100}=1.25 g$

1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr

3) 500 g of 1.25% (w/w) glucose

Mass of solution = 500 g

Mass of solute = x

Required w/w % of solution = 1.25%

$1.25\%=\frac{x}{500 g}\times 100$

$x=\frac{1.25\times 500 g}{100}=6.25 g$

6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)

4) 750 g of 2.00% (w/w) sulfuric acid.

Mass of solution = 750 g

Mass of solute = x

Required w/w % of solution = 2.00%

$2.00\%=\frac{x}{750 g}\times 100$

$x=\frac{2.00\times 750 g}{100}=15.0 g$

15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.

3 0

3 years ago

List the six steps of one full cycle of the scientific method post-lab

STatiana [176]

Ask a question
research
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7 0

3 years ago