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zloy xaker [14]
3 years ago
8

Difference between desirable changes and undesirable changes​

Chemistry
1 answer:
kkurt [141]3 years ago
3 0

Answer:

...n..m..mm .......mmnjjjjjjj ygttgyujjunjhh

Explanation:

ghyhhhhjn

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Would a false positive from the reaction between the inoculating loop and hydrogen peroxide
Svetradugi [14.3K]
The false positive from the response of hydrogen peroxide and the immunizing circle would be created by poor specificity. The recipe for specificity is TN/TN+FP. False-positive outcomes can be ascribed to meddling substances in nature where the strips are put away or utilized, for example, hydrogen peroxide (H2O2) or fade (hypochlorite).
3 0
4 years ago
I need help. Will mark Brainlist.
melomori [17]

Answer:

you are correct, the answer is 2 g

3 0
3 years ago
Select all that apply.
julia-pushkina [17]
Answer is:increase [Cl₂] and remove HCl from the product.
Chemical reaction: Cl₂ + CH₂Cl₂ → CHCl₃(chloroform) + HCl.
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7 0
4 years ago
A certain heat engine operates between 800 K and 300 K. (a) What is the maximum efficiency of the engine? (b) Calculate the maxi
Sholpan [36]

Answer :

(a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

Explanation : Given,

Temperature of hot body T_h = 800 K

Temperature of cold body T_c = 300 K

(a) First we have to calculate the maximum efficiency of the engine.

Formula used for efficiency of the engine.

\eta =1-\frac{T_c}{T_h}

Now put all the given values in this formula, we get :

\eta =1-\frac{300K}{800K}

\eta =0.625\times 100=62.5\%

(b) Now we have to calculate the maximum work done.

Formula used :

\eta =\frac{Q_h-Q_c}{Q_h}=\frac{w}{Q_h}

where,

Q_h = heat supplied by hot source = 1 KJ

Q_c = heat supplied by hot source

w = work done = ?

Now put all the given values in this formula, we get :

\eta =\frac{w}{Q_h}

0.625=\frac{w}{1KJ}

w=0.625KJ

(c) Now we have to calculate the heat discharge into the cold sink.

Formula used :

w=Q_h-Q_c

Q_c=Q_h-w

Q_c=1-0.625

Q_c=0.375KJ

Therefore, (a) The maximum efficiency of the engine is, 62.5 %

(b) The maximum work done is, 0.625 KJ.

(c) The heat discharge into the cold sink is, 0.375 KJ.

4 0
4 years ago
Ced
Aleks [24]

Answer:

c

Explanation:

7 0
3 years ago
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