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2 years ago

The further away from the sun a planet is, the hotter it is. True False

Chemistry

2 answers:

a_sh-v [17]2 years ago

7 0

Answer:

False

Explanation:

The farther away a planet is, the colder it is.

Nina [5.8K]2 years ago

3 0

the anwser is true

Explanation:

i don't like to really explain but that's the anwser it relies on its orbit

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What volume of air is needed to burn completely 100cm of propane?

Evgesh-ka [11]

Answer:

See below

Explanation:

propane mole weight = 44 gm / mole

100 gm / 44 gm / mole = 2.27 moles

From the equation, 5 times as many moles of OXYGEN (O2)are required

= 11.36 moles of oxygen

at STP this is 254.55 liters of O2 (because 22.4 L = one mole) and

Using oxygen as 21 percent of air means that

.21 x = 254.55 = x = 1212.12 liters of air required

5 0

2 years ago

Please Help 8-9!!!!!!!

seropon [69]

The labeled diagram is given in the image attached.

As it can be seen from the image that freezing is when energy is removed from the system at 0 ⁰ while melting is when energy is added at 0⁰.

Also when energy is added at 100⁰C, it causes boiling while when it is removed at 100⁰C, it causes condensation.

Melting point of water is 0⁰C while boiling point is 100⁰C

3 0

3 years ago

Drag the item from the idea bank to its corresponding match

babymother [125]

Hope this is helpful

5 0

3 years ago

Determine the formula weight of BeCl₂.

tia_tia [17]

Be= Beryllium= weight 9.01
Cl= Chlorine= weight 35.45
Since there are 2 Chlorine, you have to add Chlorine twice.
9.01 + 35.45 + 35.45= 79.91

Answer: 79.91
(This answer does not include sig.figs)

4 0

3 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

Thus, percentage abundance of 121 Sb is = 57.2 %

percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %

6 0

3 years ago