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OLga [1]
3 years ago
15

Which method is best for separating most of a solid from a gas quickly?

Chemistry
2 answers:
klemol [59]3 years ago
8 0

Answer:

I think the answer is distillation

Ivahew [28]3 years ago
5 0

Answer:

Filtration.

Explanation:

You might be interested in
Will a horse galloping in a field produce and average velocity of zero
motikmotik

Answer:

No

Explanation:

Depending on the mass of the horse and the speed, velocity will change.

5 0
3 years ago
A 10-gram aluminum cube absorbs 677 joules when its temperature is increased from 50°C to 125°C. What is the specific heat of al
ra1l [238]
<h3>Answer:</h3>

0.90J/g°C

<h3>Explanation:</h3>

We are given:

Mass of Aluminium = 10 g

Quantity of heat = 677 Joules

Change in temperature = 125°C - 50°C

                                      = 75°C

We are required to calculate the specific heat capacity of Aluminium

But, Quantity of heat = Mass × specific heat × Change in temperature

Q = mcΔt

Rearranging the formula;

c = Q ÷ mΔt

  = 677 J ÷ (10 g × 75°C)

  = 677 J ÷ 750g°C

  = 0.903 J/g°C

  = 0.90J/g°C

Thus, the specific heat capacity of Aluminium is 0.90J/g°C

8 0
3 years ago
Which is a type of star system?
Pepsi [2]

<h2><em>hope my answer is useful..</em></h2><h2><em>hope my answer is useful.. </em></h2>

7 0
1 year ago
34.969amu)(0.7577) =<br> (36.966amu)(0.2423)
Natasha_Volkova [10]

Answer:

26.4960 is the answer for the first one

8.9569 is the answer for the second one

3 0
3 years ago
Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what
Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0
3 years ago
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