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FromTheMoon [43]
3 years ago
8

Sodium hydroxide dissolves in water ....

Chemistry
1 answer:
kumpel [21]3 years ago
3 0

n the reaction will be Exothermic, where heat will be released. .

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A buret is filled with 0.1517 M A 25.0 mL portion of an unknown acid and two drops of indicator are added to an Erlenmeyer flask
erik [133]

Answer:

Molarity of Unknown Acid = 0.1332 M

Explanation:

Data for solving problem:

Molarity of base in buret (M₁)= 0.1517 M

volume of the acid in Erlenmeyer flask (V₂)= 25.0 mL

Volume of the base in the buret (V₁) = final volume of buret - initial volume in buret

final volume of buret = 22.5 mL

initial volume in buret = 0.55 mL

So

Volume of the base in the buret (V₁) = 22.5 mL -0.55 mL = 21.95 mL

Volume of the base in the buret (V₁)  = 21.95 mL

Molarity of Unknown acid in the Erlenmeyer flask (M₂) = To be find

Explanation:

It is acid base titration and  formula for this titration is as follows:

Molarity of base x Volume of base = Molarity of acid x volume of acid

it can be written as

M₁V₁ = M₂V₂ -------------------- equation (1)

we have to find M₂

so by rearrangment the equation (1)

M₁V₁ / V₂ = M₂ ------------------ equation (2)

put the values in equation in equation (2)

M₂ = 0.1517 M x 21.95 mL / 25.0 mL

M₂ = 3.3298 /25.0

M₂ = 0.1332 M

so the Molarity of Unknown acid is <u>0.1332 M</u>

5 0
3 years ago
Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

8 0
3 years ago
Carl crumbled a piece of paper before throwing it in the trash can. Which is true about the paper after it has been crumbled
tiny-mole [99]
Answer is <span>C: It's appearance changed.</span>
4 0
2 years ago
What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 40.0 minutes? Assume first-order
yaroslaw [1]

Answer:

See below

Explanation:

.75 = 1/2^(40/h)      

log .75 / ( log 1/2) = 40 / h

<u>h = half life =   96.37683 min</u>

8 0
1 year ago
List two detection (i.e. visualisation) techniques commonly used to visualise<br> compounds in TLC.
olya-2409 [2.1K]

Answer:

The most common non-destructive visualization method for TLC plates is ultraviolet (UV) light. A UV lamp can be used to shine either short-waved (254nm) or long-waved (365nm) ultraviolet light on a TLC plate with the touch of a button

Explanation:

hope this helps

7 0
3 years ago
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