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LenKa [72]
3 years ago
5

One job of the kidneys is to A. produce urea from amino groups and ammonia. B. destroy old red blood cells. C. regulate the pH o

f the blood. D. increase the salt vs. water balance in the blood.
Chemistry
1 answer:
ira [324]3 years ago
3 0
I believe the answer is A.
The kidneys will control the pH of blood to help keep the body functioning properly. 
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8.5C
DENIUS [597]

Answer:

i.e belongs to same group because of valence electrons are same

Explanation:

so it has same chemical behaviour. and q has more energy than r ionisation energy decreases from top to bottom

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3 years ago
If a system requires 150 J of input work and produces 123 J of output work, what's its efficiency? A. 72% B. 122% C. 82% D. 92%
STALIN [3.7K]
The answer is C. 82......
4 0
3 years ago
Read 2 more answers
What is a molecular formula?
polet [3.4K]
The symbolic representation of its compound and its composition. 

or a chemical formula that indicates the kinds of atoms and the number of each kind in a molecule of a compound. 
ANSWER IS B
5 0
3 years ago
Which of the following is non polar covalent bond?
vladimir2022 [97]

Answer:

N-Cl

Explanation:

Look at the chart below. Since N-Cl bond has a electronegativity difference of (3.0-3.0) zero, they are non-polar.

5 0
2 years ago
Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)
Fynjy0 [20]

Answer:

Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
3 years ago
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