Answer:
its c
Explanation:
solvent is the part of the solution that is in the greater quantity, and solute is the smaller portion of the solution
hope this helps :)
I believe that the number for Co2 in 220 grams is 5
Answer:
The answer is 0.36 kg/s NO
Explanation:
the chemical reaction of NH3 to NO is as follows:
4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)
We have the following data:
O2 Volume rate = 645 L/s
P = 0.88 atm
T = 195°C + 273 = 468 K
NO molecular weight = 30.01 g/mol
we calculate the moles found in 645 L of O2:
P*V = n*R*T
n = P*V/R*T
n= (0.88 atm * 645L/s)/((0.08205 L*atm/K*mol) * 468 K) = 14.78 moles of O2
With the reaction we can calculate the number of moles of NO and with its molecular weight we will have the rate of NO:
14.78 moles/s O2 * 4 molesNO/5 molesO2 * 30.01 g NO/1 molNO x 1 kgNO/1000 gNO = 0.36 kg/s NO
There are four types of chemical bonds essential for life to exist: Ionic Bonds, Covalent Bonds, Hydrogen Bonds, and van der Waals interactions. We need all of these different kinds of bonds to play various roles in biochemical interactions. These bonds vary in their strengths.
To play a variety of roles in biochemical interactions, we require all of these diverse sorts of linkages. The tensile strength of these linkages varies. In chemistry, we consider the range of strengths between ionic and covalent bonds to be overlapping. This indicates that in water, ionic bonds usually dissociate. As a result, we shall consider these bonds from strongest to weakest in the following order:
Covalent is followed by ionic, hydrogen, and van der Waals.
To know more about 4 different types of bonds, visit;
brainly.com/question/17401243
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Given :
2NOBr(g) - -> 2NO(g) + Br2(g)
Initial pressure of NOBr , 1 atm .
At equilibrium, the partial pressure of NOBr is 0.82 atm.
To Find :
The equilibrium constant for the reaction .
Solution :
2NOBr(g) - -> 2NO(g) + Br2(g)
t=0 s 1 atm 0 0
1( 1-2x) 2x x
So ,

At equilibrium :
![K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cdfrac%7B%5BNO%5D%5E2%5Bbr_2%5D%7D%7B%5BNOBr%5D%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D%5Cdfrac%7B0.18%5E2%5Ctimes%200.9%7D%7B0.82%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D0.043%5C%20atm)
Hence , this is the required solution .