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Over [174]
3 years ago
5

Please help both these answers use the graph . I will mark brainly !!

Chemistry
1 answer:
____ [38]3 years ago
3 0

1 is 4

2 is d

hope it helps


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You have a graduated cylinder with 10 mL of water in it.
Law Incorporation [45]

Answer:

<h3>The answer is 10 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 300 g

volume = final volume of water - initial volume of water

volume = 40 - 10 = 30 mL

We have

density =  \frac{300}{30}  =    \frac{30}{3}  \\

We have the final answer as

<h3>10 g/mL</h3>

Hope this helps you

5 0
3 years ago
Alveoli are tiny sacs of air in the lungs. Their average diameter is 4.50 × 10−5 m. Calculate the uncertainty in the velocity of
Arisa [49]

<u>Answer:</u> The uncertainty in the velocity of oxygen molecule is 4.424\times 10^{-5}m/s

<u>Explanation:</u>

The diameter of the molecule will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

\Delta x.\Delta p=\frac{h}{2\pi}

where,

\Delta x = uncertainty in position = d = 4.50\times 10^{-5}m

\Delta p = uncertainty in momentum  = m\Delta v

m = mass of oxygen molecule = 5.30\times 10^{-26}kg

h = Planck's constant = 6.627\times 10^{-34}kgm^2/s^2

Putting values in above equation, we get:

4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 4.50\times 10^{-5}m\times 5.30\times 10^{-26}kg}=4.424\times 10^{-5}m/s

Hence, the uncertainty in the velocity of oxygen molecule is 4.424\times 10^{-5}m/s

8 0
3 years ago
Someone help me am i right or wrong
steposvetlana [31]

Answer:

You right!

Explanation:

3 0
3 years ago
Read 2 more answers
Which object has stored gravitational potential energy
kakasveta [241]

Answer:a lightbulb burning

Explanation:

Apex

5 0
3 years ago
Read 2 more answers
The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M
anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0
2 years ago
Read 2 more answers
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