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olya-2409 [2.1K]
3 years ago
14

Describe general characteristics of potassium

Chemistry
1 answer:
Virty [35]3 years ago
6 0
Leg crams are symptoms of lack of potassium
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Ammonia and oxygen react to form nitrogen monoxide and water, like this:
olchik [2.2K]

4NH3+5O2 <=>4NO + 6H2O

Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.

The numerical value for a given temperature can be evaluated when the actual partial pressures are known.
6 0
4 years ago
What is the molarity of a solution if you take 12.0 grams of ca(no3)2 and mix it with 0.105 l of water?
Lunna [17]
The  molarity  of  a solution  if it tale 12.0   grams  of  Ca(No3)2  is  calculated as below

molarity =  moles/volume in liters
 moles = mass/molar mass =  12.0 g/ 164 g/mol =  0.073 moles

molarity is therefore  = 0.073/0.105 = 0.7 M 
8 0
3 years ago
A screw is an example of a modified 1 pulley 2 Wedge 3 inclined plane 4 wheel​
valentinak56 [21]

Answer: The answer would be D.

Explanation: When a screw is tightened into an object, it is wedge in between.

8 0
2 years ago
Adding energy to a solid bar of gold may result in which of the following outcomes
____ [38]
An increase in motion and less attraction between particles

4 0
3 years ago
A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur
lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

q=m\times c\times \Delta T

q = heat absorbed by water

m = mass of water = 40.0 g

T_{final} = final temperature of water = 20.0°C

T_{initial = initial temperature of water = 17.0°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

q=40.0\times 4.186\times (20.0-17.0)]

q=502J

Hence, the joules of heat were re-leased by the lead is 502

5 0
3 years ago
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