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Ainat [17]
2 years ago
9

2. Compare Which is NOT a good place to look for life?

Chemistry
1 answer:
Verdich [7]2 years ago
4 0

The Sun

Explanation:

Because it's to hot to live on

You might be interested in
Metals present in municipal wastewater may still be present in treated sewage sludge; ______
Marrrta [24]
Metals present in municipal waste water may still be present in treated sewage sludge IN CONCENTRATIONS THAT MAY AFFECT THE PUBLIC HEALTH. Sewage sludge is an end product of municipal waste water treatment and it contains many of the pollutant that are removed from the waste water. 
5 0
3 years ago
Which two compounds are classified as bases by the Brønsted-Lowry definition, but not by the Arrhenius definition, and why?
konstantin123 [22]

Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.

Explanation:

i took the test and got it correct :) hope this helps

6 0
3 years ago
7. What is the molarity, M, of the dilution solution of KBr, given:
sineoko [7]

Answer:

Hello There!!

Explanation:

I believe the answer is c. Volume of the dilute solution is 100.00 mL.

hope this helps,have a great day!!

~Pinky~

7 0
3 years ago
Read 2 more answers
Calculate the mass of KHC8H4O4 that reacts with 20 ml of 0.1 Monday NaOH.
Zigmanuir [339]
 <span>Obviously only one of the H's was replaced as the product showed one mole of water So the ratio is 1 to 1.Moles of NaOH used Molarity x volume = 0.1 x0.015(L) = 0.0015 moles 
Mass of acid used is 0.0015 x 204 = 0.306g 
moles MM

hopefully this helps</span>
3 0
3 years ago
What is the pressure of 0.33 moles of nitrogen gas, if its volume is 15.0 L at –25.0oC?
musickatia [10]

Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=

V

nRT

​

 . Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}

mol⋅K

atm⋅L

​

 :

P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=

V

nRT

​

=

15.0

L

​

(0.33

mol

)(0.0821

mol⋅K

atm⋅

L

​

​

)(248.15

K

​

)

​

=0.4482atm

In conclusion, the pressure of this gas is P=0.4482 atm.

Reference:

Chang, R. (2010). Chemistry. McGraw-Hill, New York.

3 0
2 years ago
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