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Sedaia [141]
3 years ago
8

What products will I get if I add potassium metal to water?​

Chemistry
1 answer:
docker41 [41]3 years ago
8 0
2K+2H2O=>2KOH+H2.
The reaction is exothermic.
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Which of the following represents a chemical change?
Snezhnost [94]
The answer is B.) Freezing of water
3 0
3 years ago
if a gas sample in a balloon occupies 1.5 L at atmospheric pressure, what would be the pressure (in mmHg) if the volume was redu
Nata [24]

Answer:

1425 mmHg.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 1.5 L

Initial pressure (P1) = 1 atm

Final volume (V2) = 0.8 L

Final pressure (P2) =?

Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:

P1V1 = P2V2

1 × 1.5 = P2 × 0.8

1.5 = P2 × 0.8

Divide both side by 0.8

P2 = 1.5/0.8

P2 = 1.875 atm

Finally, we shall convert 1.875 atm to mmHg.

This can be obtained as follow:

1 atm = 760 mmHg

Therefore,

1.875 atm = 1.875 × 760 = 1425 mmHg.

Therefore, the new pressure of the gas is 1425 mmHg.

3 0
3 years ago
The equation K + NaOH => Na + KOH should it react, true or false?
Andru [333]

Answer:

yes it should react yoyoyoyoyo

3 0
2 years ago
If 1.76 g of an ideal gas occupy 1.0 L at standard temperature and pressure (STP), what is the molar mass of the gas?
ycow [4]

Answer:

Explanation:

Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation PV=nRT to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:M=dRTPM = molar mass (g/mol)d = density (g/L)R = Ideal Gas Constant (≈0.0821atm⋅Lmol⋅K) T = Temperature (In Kelvin) P = Pressure (atm)As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of. Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:T=35oC+273.15= 308.15 KV=300mL⋅1000mL1L= 0.300 LP=789mmHg⋅1atm760mmHg= 1.038 atmSo, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:d=0.622g0.300L= 2.073 g/LNow, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.M=dRTP=(2.073)(0.0821)(308.15)1.038= 51 g/molRounded to 2 significant figuresNow if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). Hope that helped :) 

8 0
2 years ago
The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a
Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

Mass remain after 49.2 years = ?

Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives =  49.2 years /12.3 years

Number of half lives =  4

Now we will calculate the amount left after 49.2 years.

At time zero 48.0 mg

At first half life = 48.0mg/2 = 24 mg

At second half life = 24mg/2 = 12 mg

At 3rd half life = 12 mg/2 = 6 mg

At 4th half life =  6mg/2 = 3mg

The amount left after 49.2 years is 3mg.

6 0
3 years ago
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