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3 years ago

9

The properties of a substance are not affected by chemical reactions. O True O False

1 answer:

ElenaW [278]3 years ago

3 0

Answer:

False

Explanation:

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Please help I will mark Brainly!

IgorLugansk [536]

It's NF3 (one Nitrogen and 3 Flouride)

6 0

3 years ago

We dissolve 93 grams of sodium sulfate (Na2SO4) in 20 grams of water. What is the percent by mass of sodium sulfate in this solu

dlinn [17]

It should be for the total solution of 93 plus 20 grams which is 113 grams so 93 divided by 113 grams comes to 82.3% sodium sulfate and this can be checked by multiplying 113 grams by 0.823 which results in 93 grams of sodium sulphate.

4 0

4 years ago

A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume

lbvjy [14]

We have that every gas satisfies the fundamental gas equation, PV=nRT where P is the Pressure, V is the volume of the gas, n are the moles of the gas, R is a universal constant and T is the Temperature in Kelvin. We have that PV/T=nR and during our process, the moles of the gas do not change (no argon enters or escapes our sample). See attached.

6 0

4 years ago

A certain first-order reaction 45% complete in 65seconds, determine the rate constant and the half life for the process

masha68 [24]

The rate constant : k = 9.2 x 10⁻³ s⁻¹

The half life : t1/2 = 75.3 s

<h3>Further explanation</h3>

Given

Reaction 45% complete in 65 s

Required

The rate constant and the half life

Solution

For first order ln[A]=−kt+ln[A]o

45% complete, 55% remains

A = 0.55

Ao = 1

Input the value :

ln A = -kt + ln Ao

ln 0.55 = -k.65 + ln 1

-0.598=-k.65

k = 9.2 x 10⁻³ s⁻¹

The half life :

t1/2 = (ln 2) / k

t1/2 = 0.693 : 9.2 x 10⁻³

t1/2 = 75.3 s

3 0

3 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

3 years ago