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3 years ago

Express the rate of reaction in terms of the change in concentration of each of the reactants and products. 2A(g) → B(g) + C(g)

Chemistry

1 answer:

ankoles [38]3 years ago

8 0

[B][C] / [A]^2

Products raised to the coefficients over reactants raised to the coefficients

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Give the chemical symbols for the following elements: (a) potassium, (b) tin, (c) chromium, (d) boron,(e) barium, (f) plutonium,

Ilia_Sergeevich [38]

Answer:

A) Alkili Metal

B) Basic Metal

C) Transition Metal

D) Semimetal

E) Alkiline Earth

F) Actinide

G) Nonmetal

H) Noble Gas

I) Transitional Metal

Hope This Helps

8 0

3 years ago

Suggest two reasons why the noble gases become less ideal in their behavior down the group from helium to xenon.

katrin2010 [14]

Here is your answer !!! Last day we have solve this question and the answer is easy to get in mind

6 0

3 years ago

A car with a mass of 1,200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. The centri

Helga [31]

The answer is 80 m.

Centripetal force (F) is a force that makes body move around a circular curve. The unit of force is N (N = kg * m/s²).
It can be represented as:
$F= \frac{m* v^{2} }{r}$
where:
m - mass
v - velocity
r - radius of the curve

We have:
m = 1,200 kg
V = 20 m/s
F = 6,000 N = 6,000 kg * m/s²

We need radius of the curve:
r = ?

So, if $F= \frac{m* v^{2} }{r}$ , then:
 $r= \frac{3* v^{2} }{F}$
⇒ $r = \frac{1200 kg * (20m/s)^{2} }{6000 kg*m/s^{2} }$
⇒ $r= \frac{1200*400*kg*m^{2}*s^{2} }{6000kg*m/s^{2} }$
⇒ $r= \frac{480,000m}{6000}$
⇒ $r=80m$

6 0

3 years ago

Tap water at room temperature has a pH of 7.2 and carbonate alkalinity of 200 mg/L (= 40 mmol/L). What is the concentration of b

UNO [17]

Answer:

the concentration of bicarbonate is [HCO₃⁻] = 0,03996 M and carbonate is [CO₃²⁻] = 3,56x10⁻⁵ M.

Explanation:

Carbonate-bicarbonate is:

HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25

Using Henderson-Hasselbalach formula:

pH = pka + log₁₀ $\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}$

7,2 = 10,25 + log₁₀ $\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}$

8,91x10⁻⁴ = $\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}$ (1)

Also:

0,040 M = [CO₃²⁻] + [HCO₃⁻] (2)

Replacing (2) in 1:

[HCO₃⁻] = 0,03996 M

Thus:

[CO₃²⁻] = 3,56x10⁻⁵ M

I hope it helps.

4 0

3 years ago

Where does o2 and CO2 gas exchange occur?

mixer [17]

The O₂ and the CO₂ gas exchange occurs in alveoli. the oxygen gas moves from the lungs and at same time carbon dioxide passe the the lungs from the blood and the gas exchange occurs.

The oxygen gas moves from the lungs and at same time carbon dioxide passe the the lungs from the blood and gas exchange between the alveoli and the tiny blood vessels occurs. the tiny blood vessels are called as the capillaries. the capillaries are located at the alveoli walls. every person has the hundred of millions of the alveoli present in their lungs.

Thus, in the alveoli the exchange of the oxygen gas and the carbon dioxide gas will occurs.

To learn more about the alveoli here

brainly.com/question/6748872

#SPJ4

6 0

1 year ago