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katen-ka-za [31]
3 years ago
12

How many moles of potassium chloride (KCI) are in 252 mL of 0.33 mil/L KCI

Chemistry
1 answer:
Rom4ik [11]3 years ago
7 0

Answer:

0.083 moles KCl

Explanation:

We have 252 mL of 0.33 M KCl. "M" is molarity, and it's measured in mol/L. Since we know the molarity is 0.33 mol/L, in order to find how many moles 252 mL is, we need to multiply our given volume by the molarity.

However, our given volume is in mL, so let's convert to L first by dividing 252 by 1000 (because there are 1000 mL in 1 L):

252/1000 = 0.252 L

Now, we can multiply 0.252 L by 0.33 mol/L to get moles since the L units will cancel out:

0.252 L * 0.33 mol/L = 0.08316

We have two significant figures here, so round 0.08316 to 0.083.

The answer is 0.083 moles.

Hope this helps!

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3 years ago
Xenon is an element similar to Neon. Its density can be found on the Reference Table. Convert Xenon’s density to scientific nota
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The Scientific Notation of  Xenon’s density : 1.31293 × 10²

<h3>Further explanation</h3>

Scientific Notation consists of 2 parts

1. the first part: the first digit followed by the decimal point

2. The second part: the power of 10 which shows the number of digits after the decimal point

Xenon's density = 131.293

First digit = 1, then we give the decimal point and the next digit

The power of 10 is determined from the number of decimal points moved from the original decimal point

Because we move the decimal point two digits to the left, the power of 10 has the number 2 (10²)

So that the Scientific Notation form becomes:

1.31293 × 10²

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3 years ago
In a chemical reaction, mass is conserved. This means that
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D. the mass and number of atoms and the elements will always and must be equal on both sides of the equation, mass can not be destroyed, only conserved
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2 years ago
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically writ
Alex17521 [72]

Answer:

T = 42.08  °C

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, E_a = 45.6 kJ/mol = 45600 J/mol (As 1 kJ = 1000 J)

k_2=2\times k_1

k_1=0.0160s^{-1}

T_1=30\ ^0C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (30 + 273.15) K = 303.15 K  

T_1=303.15\ K

So,

\ln \dfrac{k_{1}}{2\times k_{1}} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )

\ln \dfrac{1}{2} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )

8.314\ln \left(2\right)=-45600\left(\frac{1}{303.15}-\frac{1}{T_2}\right)

8.314\ln \left(2\right)=-150.42058+\frac{45600}{T_2}

144.65775 =\frac{45600}{T_2}

T_2=\frac{45600}{144.65775}

T_2=315.23\ K

Conversion to °C as:

T(K) = T( °C) + 273.15  

So,  

315.23 = T( °C) + 273.15

<u>T = 42.08  °C</u>

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