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kiruha [24]
2 years ago
12

On a sunny day, you can sometimes see dust in the air. Explain why the dust moves.

Chemistry
1 answer:
Nata [24]2 years ago
3 0

Answer:

Explanation:

When wind passes through. the dust, the dust then also moves throughout the air, when you move and breath in a room, you create slight wind (like a fan moving to creating wind) and this slight wind moves the dust particles.

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The term Reuther Ford gave to the positively charged particles in the nucleus of an antom was
lubasha [3.4K]
The term Rutherford gave to the positively charged particles in the nucleus of an atom was/is Proton. 

Hope this helps!
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4 years ago
A fish has armored plates that enable it to handle the extreme pressure in the deepest, darkest aquatic ecosystem.
love history [14]

Answer:

the open ocean

Explanation:

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6 0
4 years ago
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A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate
katrin2010 [14]

Answer:

Explanation:

The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT

ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution

Kb for water = .51°C /m

moles of sugar = 16.90 / 342.3

= .04937 moles

m = moles of sugar /  kg of water

= .04937 / .04090

= 1.207

ΔT = Kb x m

= .51 x 1.207

= .62°C .

So , boiling point of water = 100.62°C .

8 0
3 years ago
What is parthenolide
vovangra [49]
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7 0
3 years ago
The initial temperature of the water in a constant-pressure calorimeter is
zubka84 [21]

Answer: The enthalpy change during this reaction is 77505.56 J.

Explanation:

Given: T_{1} = 14^{o}C,         T_{2} = 87^{o}C

Mass = 254 g,       Specific heat = 4.18 J/g^{o}C

Formula used to calculate the enthalpy change is as follows.

q = m \times C \times (T_{2} - T_{1})

where,

q = enthalpy change

m = mass of substance

C = specific heat capacity

T_{1} = initial temperature

T_{2} = final temperature

Substitute the values into above formula as follows.

q = m \times C \times (T_{2} - T_{1})\\= 254 g \times 4.18 J/g^{o}C \times (87 - 14)^{o}C\\= 77505.56 J

Thus, we can conclude that the enthalpy change during this reaction is 77505.56 J.

3 0
3 years ago
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