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zavuch27 [327]
3 years ago
14

Why can't air exist in space, but gas can????

Chemistry
1 answer:
MakcuM [25]3 years ago
8 0

Answer:

Earth's gravity is strong enough to hold onto its atmosphere and keep it from drifting into space.

You might be interested in
Which method is used to obtain petrol from petroleum?
kotykmax [81]

Explanation:

fractional distillation method is used to obtain petrol from petroleum...

hope it helps

6 0
3 years ago
Read 2 more answers
If 335 g water at 35.5C loses 5750 J of heat, what is the final temperature of the water?
White raven [17]

Answer:

The final temperature is  39.58 degree Celsius

Explanation:

As we know

Q = m * c * change in temperature

Specific heat of water (c) = 4.2 joules per gram per Celsius degree

Substituting the given values we get  -

5750 = 335 * 4.2 * (X - 35.5)

X = 39.58 degree Celsius

3 0
3 years ago
What needs to happen for the equation CH4 + O2 → CO2 + H2O to be balanced?
worty [1.4K]
B. This is because the Hydrogen and Oxygen need balanced out.

Current-

C-1            |        C-1
H-4            |         H-2
O-2            |         O-3


Adding a coefficient of 2 before oxygen in the reactants and H2O in the products would balance this equation

<span>CH4 + 2O2 → CO2 + 2H2O</span>
C-1            |        C-1
H-4            |         H-4
O-4            |        O-4


4 0
3 years ago
In the space provided, for each element, type the ionic charge of an atom with a full set of valence electrons. Then, type the n
kobusy [5.1K]

11. ionic charge +1, helium.

12. ionic charge 2-, neon.

13. ionic charge 3+, neon.

3 0
3 years ago
Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.
TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

HCl+KOH\rightarrow KCl+H_2O

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

n_{KOH}=0.00768mol-0.00192mol=0.00576mol

And the resulting concentration of KOH and OH ions as this is a strong base:

[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M

And the resulting pH is:

pH=14+log(0.131)\\\\pH=13.1

Regards!

3 0
3 years ago
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