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sergij07 [2.7K]
3 years ago
8

When a customer begins choking you should administer CPR. True False

Chemistry
2 answers:
nydimaria [60]3 years ago
6 0
True because it’s just common sense
yuradex [85]3 years ago
3 0

Answer:

True

Explanation:

That's my answer :)

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PLEASE I NEED YOUR HELP, I WILL GIVE 10 POINTS PLEASEEEEE
Cerrena [4.2K]

Answer:

Erosion

Explanation:

3 0
3 years ago
Exactly one mole of an ideal gas is contained in a 2.00 liter container at 1,000 K. What is the pressure exerted by this gas?
Kipish [7]
pV = nRT

p = nRT/V 

p= 1 x 0.08205 x 1000/ 2

p = 41.025 Pa  

Edit: The unit should be atm instead of Pa, as pointed out by a nice human being.
4 0
4 years ago
Read 2 more answers
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode wh
morpeh [17]

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

8 0
3 years ago
What is the nucleus of an atom split
larisa [96]
In nuclear physics and nuclear chemistry, nuclear fission is either a nuclear reaction or a radioactive decay process in which the nucleus of an atom splits<span> into smaller parts (lighter </span>nuclei<span>). Hope this helps</span>
4 0
3 years ago
Read 2 more answers
For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper
Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1\text{ and }V_1 are initial pressure and volume at initial temperature T_1.

P_2\text{ and }V_2 are final pressure and volume at initial temperature T_2.

We are given:

P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V

T_1=-25^oC=248.15 K

T _2 = 16^oC=289.15 K

Putting values in above equation, we get:

\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}

P_1=0.225 kPa

Hence, the new pressure will be 0.225  kPa.

4 0
4 years ago
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