We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, 
moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
0.25 moles of CO2 is present in 11 grams of CO2.
Explanation:
A mole represents the number of chemical entities in an element or molecule.
Number of moles of an element or molecule is determined by the formula:
The Number of moles (n) = weight of the atom given ÷ atomic or molecular weight of the one mole of the element or molecule.
Themolar mass of one mole of carbon dioxide is:
12+ ( 16×2)
= 44 gram/mole
The given weight is 44 grams of carbon dioxide.
Putting the values in the equation,
n= 11 gms÷44 gms/ mole
n = 0.25 mole
Explanation:
( a )
<u>The four types of spread spectrum techniques are as follows -</u>
1. Direct sequence spread spectrum .
2. frequency hopping spread spectrum .
3. chirp spread spectrum .
4. time hopping spread spectrum .
( b )
<u>The Direct sequence spread spectrum was devised for eavesdropping in the military .</u>
In the field of telecommunications , the Direct sequence spread spectrum , it is the technique of spread spectrum modulation which is used to reduce the overall inference of the signal .
<h3>
Answer:</h3>
1379.4 Joules
<h3>
Explanation:</h3>
- The quantity of heat is calculated multiplying the mass of a substance by heat capacity and the change in temperature.
Therefore;
Quantity of heat = Mass × specific heat capacity × Change in temperature
Q = mcΔT
In this case;
The substance dissolved in water gained heat while water lost heat energy.
Thus, Heat gained by the substance = heat lost by water
Heat associated with the water
Mass of water = 75 g
Change in temperature = 4.4°C
Specific heat capacity = 4.18 J/g·⁰C
Heat = mcΔT
= 75 g × 4.18 J/g·⁰C × 4.4 °C
=1379.4 Joules
Answer:
pH = 6.999
The solution is acidic.
Explanation:
HBr is a strong acid, a very strong one.
In water, this acid is totally dissociated.
HBr + H₂O → H₃O⁺ + Br⁻
We can think pH, as - log 7.75×10⁻¹² but this is 11.1
acid pH can't never be higher than 7.
We apply the charge balance:
[H⁺] = [Br⁻] + [OH⁻]
All the protons come from the bromide and the OH⁻ that come from water.
We can also think [OH⁻] = Kw / [H⁺] so:
[H⁺] = [Br⁻] + Kw / [H⁺]
Now, our unknown is [H⁺]
[H⁺] = 7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]
[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) / [H⁺]
This is quadratic equation: [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴
a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴
(-b +- √(b² - 4ac) / (2a)
[H⁺] = 1.000038751×10⁻⁷
- log [H⁺] = pH → 6.999
A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.
