The nitrate reduction test identifies whether the given strain of bacteria is able to reduce nitrate to nitrite using anaerobic respiration.
In this test, the solution is incubated after adding the bacteria strain. After the incubation period is over, additional solutions Sulfuric acid (a) and Naphthylamine (b) are added to it.
If the reduction has occurred and the nitrite is present, then the solution gives red color.
If there is no color change, then the solution is tested for the presence of other reduction products like N2 gas , NH3 gas etc. This is done by adding zinc to the solution.
Zinc is able to reduce nitrate to nitrite only. Therefore a color change after the addition of zinc indicates that the bacterial strain was not able to reduce nitrate and nitrates are still present which were reduced by zinc.
If there is no color change, that means the nitrates are absent and the bacteria has reduced nitrate to some other species like N2 or NH3 gas.
Therefore , In the nitrate reduction test, only after the addition of zinc to your 24-hour incubated solutions a and b is there a color change. This indicates that the bacteria was not able to reduce nitrate to nitrite.
Plastic or rubber as non metal are bad conductor of electricity.
Answer: +35
explanation: 30+35£~£
Answer:
Explanation:
When Mg reacts with HCl, magnesium chloride and hydrogen is formed. Mg is an active element and displaces hydrogen from HCl. So, this is a type of single displacement reaction.
When magnesium oxide (MgO) reacts with HCl, magnesium chloride and water is formed. This reaction is a type of neutralization reaction. MgO is a water insoluble base and HCl is acid. So. in this reaction, acid reacts with base to form salt 
Answer:
73.49 atm
Explanation:
The balance chemical equation is as follow,
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Given Data:
Temperature = T = 373 K
Volume = V = 2.5 L
Gas Constant = R = 0.0821 atm.L.mol⁻¹.K⁻¹
To find:
Pressure: P = ?
Formula:
P V = n R T
P = n R T / V
Here we also need to find number of mole,
According to balanced equation,
1 mole of C₃H₈ on combustion gives = 3 moles of CO₂
So,
2 moles of C₃H₈ on combustion will give = X moles of CO₂
Solving for X,
X = 2 moles of C₃H₈ × 3 moles of CO₂ ÷ 1 mole of C₃H₈
X = 6 moles of CO₂
Moles = n = 6 mol
Solution:
P = 6 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 373 K ÷ 2.5 L
P = 73.49 atm
