Answer:
a) No. of moles of hydrogen needed = 5.4 mol
b) Grams of ammonia produced = 27.2 g
Explanation:
a)
No. of moles of nitrogen = 1.80 mol
1 mole of nitrogen reacts with 3 moles of hydrogen
1.80 moles of nitrogen will react with
= 1.80 × 3 = 5.4 moles of hydrogen
b)
No. of moles of hydrogen = 2.4 mol
It is given that nitrogen is present in sufficient amount.
3 moles of hydrogen produce 2 moles of 
2.4 moles of hydrogen will produce
= 
Molar mass of ammonia = 17 g/mol
Mass in gram = No. of moles × Molar mass
Mass of ammonia in g = 1.6 × 17
= 27.2 g
C because I’m space there’s no gravity meaning things are weightless
Answer: an ex ray and a fire
Explanation:
Answer : The pH of the solution is, 1.88
Explanation : Given,
Concentration of 
= 0.35 M
Concentration of 
= 0.29 M
First we have to calculate the value of 
.
The expression used for the calculation of is,
![pK_a=-\log [K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%20%5BK_a%5D)
Now put the value of 
in this expression, we get:
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[NaClO_2]}{[HClO_2]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BNaClO_2%5D%7D%7B%5BHClO_2%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of the solution is, 1.88
Answer:
1. The pressure in outer space is so low that many consider it as non-existant. It has a pressure of 1.322 × 10−11 Pa. Pressure may be detected from the molecule of air or water hitting you. Since there is very little air and hardly ever water hitting you in space, pressure is almost zero or negligible.
2. Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, 14.70 pounds per square inch, 1,013.25 × 103 dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.
3. 0.46J/gC
explanation.The specific heat capacity of a material is given by:
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C_s = \frac{Q}{m \Delta T}C s = mΔTQ
where
Q is the amount of heat absorbedm is the mass
\Delta TΔT is the variation of temperature
For the piece of iron in the problem:
m = 15.75 gm=15.75g
Q=1086.75 JQ=1086.75J
\Delta T=175 C-25 C=150^{\circ}ΔT=175C−25C=150∘
Substituting into the equation,
C_s = \frac{1086.75 J}{(15.75 g)(150^{\circ}C)}=0.46 J/gCC s = (15.75g)(150 ∘ C)1086.75J = 0.46J/gC
4. 207 J
Explanation:
The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of 1 g of a given substance by 1∘C.
#I hope it's help
