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Andrews [41]
3 years ago
5

What does a Stationary object line look like?

Chemistry
1 answer:
Paladinen [302]3 years ago
5 0

Answer:

Explanation:It looks like a graph. Google it!

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A compound which is mineral acid is generally classified as a????? urgent please​
andrey2020 [161]

Answer: An inorganic acid.

Explanation: An inorganic acid (also called a mineral acid) is an acid derived from one or more inorganic compounds. All inorganic acids form hydrogen ions and the conjugate base ions when dissolved in water. Commonly used inorganic acids are sulfuric acid (H2SO4), hydrochloric acid (HCl), and nitric acid (HNO3). Hope this helps :)

3 0
3 years ago
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what is the molarity of a RbOH solution if 60.0 mL of the solution is neutralized by 52.8 mL of a 0.5M HCl solution (Hint: Ma x
Dahasolnce [82]
RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume 
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted 
Ma x Va = Mb x Vb
0.5 M x 52.8 mL = Mb x 60.0 mL 
Mb = 0.44 M 
molarity of base - 0.44 M 

7 0
3 years ago
What do you expect would happen to the pressure of the gas in a container as the volume is decreased
Over [174]

Answer:

the pressure would increase

Explanation:

5 0
2 years ago
For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:
KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0
3 years ago
Is dynamite an atom or a molecule?
kkurt [141]

Answer:

TNT is a molecule

Explanation:

4 0
3 years ago
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