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artcher [175]
3 years ago
14

When a reaction occurs between atoms with ground state electron configurations 1s 2s' and 1s 2s 2p%, the predominant type of bon

d formed is:
Chemistry
1 answer:
Andrej [43]3 years ago
5 0
Figure 5.9 argon
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If I have 6 moles of a gas at a pressure of 3.4 atm and a volume of 10 liters, what
ch4aika [34]

Answer:

The temperature is 69.05 K.  

Explanation:

We have,

Number of moles are 6

Pressure is 3.4 atm and a volume of 10 liters.

It is required to find the temperature. It can be calculated using gas law equation. It says that,

PV=nRT

P is pressure

V is volume

n is number of moles

R is gas constant, R=0.082057\ L-atm/mol-K

T is temperature                

Plugging all the values we get :

T=\dfrac{PV}{nR}\\\\T=\dfrac{3.4\times 10}{6\times 0.082057}\\\\T=69.05\ K

So, the temperature is 69.05 K.  

8 0
3 years ago
The properties of a substance are not affected by chemical reactions.<br> O True<br> O False
ElenaW [278]

Answer:

False

Explanation:

3 0
3 years ago
Doctors have fought infections with penicillin for over 50 years. Recently, penicillin has not worked as well as before. Which o
Lilit [14]

C. "A" mentions that the drug killed all of the bacteria that are resistant to the drug, and that doesn't make any sense. "B" claims that all bacteria are resistant to the drug. This is not true. "D" mentions that bacteria eat the drug, which doesn't happen.

5 0
3 years ago
To what temperature must a balloon, initially at 25°c and 2.00 l, be heated in order to have a volume of 6.00 l? question 8 opti
cluponka [151]
<span>a.655 k not 100 percent on this but try it. You will use 273.15 and add your Celcius temp to get it in Kelvin
</span>
5 0
3 years ago
Read 2 more answers
The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process
denis23 [38]

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

7 0
3 years ago
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