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Dafna1 [17]
3 years ago
5

1-Which reaction shown below is carried out by nitrifying bacteria? I will mark you as brainlist

Chemistry
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

ammonium to nitrate.

Explanation:

they tie em together

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4 NH3 + 6 NO → 5 N2 + 6 H2O How many moles of NH3 are necessary to produce 0.824 mol N2?
GrogVix [38]
4 mol NH₃ → 5 mol N₂
x mol NH₃ → 0.824 mol N₂

x=0.824*4/5=0.6592 mol
6 0
3 years ago
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

5 0
3 years ago
What fraction of a sample of 6832ge, whose half-life is about 9 months, will remain after 2.9 yr ?
Viefleur [7K]
<span>The half-life of 9 months is 0.75 years. 2.0 years is 2.0/0.75 = 2.67 half-lives. Each half-life represents a reduction in the amount remaining by a factor of two, so: A(t)/A(0) = 2^(-t/h) where A(t) = amount at time t h = half-life in some unit t = elapsed time in the same unit A(t)/A(0) = 2^(-2.67) = 0.157 15.7% of the original amount will remain after 2.0 years. This is pretty easy one to solve. I was happy doing it.</span>
4 0
3 years ago
Binary compounds are formed by ............... ............... elements.
goblinko [34]

Answer: i think its A diatomic compound..

Explanation: hope i helped! sorry if im wrong!

6 0
3 years ago
What volume of water is required to prepare 0.1 M H3PO4 from 100 ml of 0.5 M solution?
ExtremeBDS [4]

Answer: A volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

Explanation:

Given: M_{1} = 0.1 M,    V_{1} = ?

M_{2} = 0.5 M,       V_{2} = 100 mL

Formula used to calculate the volume of water is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

7 0
3 years ago
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