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2 years ago

1. What volume of a 2.50M Kl(aq) is needed to make 200 ml of a 1.OOM KI)aq)?

Chemistry

1 answer:

FinnZ [79.3K]2 years ago

8 0

Answer:

80 ml

Explanation:

From the question,

Applying Dilution formular

MV = mv................... Equation 1

Where M = Molarity of Kl before dilution, V = Volume of Kl before dilution, m = molarity of Kl after dilution, v = volume of Kl after dilution.

make V the subject of the equation

V = mv/M............. Equation 2

Given: m = 1.00 M, v = 200 ml, M = 2.50 M

Substitute these values into equation 2

V = (1.00×200)/2.50

V = 80 ml

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A sample of propane(c3h8)has 3.84x10^24 H atoms.

deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

8 0

1 year ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

Answer: The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

2 years ago

Which of the following is not true of a scientific theory?

Furkat [3]

Answer: a,b

Explanation: because a scientific theory is something that been thought of and tested multiple times.

3 0

2 years ago

What is the engine piston displacement in liters of an engine who’s displacement is listed as 430 in.²

sergeinik [125]

For an engine whose displacement is listed as 430 in.², the engine piston displacement in liters is mathematically given as

PD= 7.37 L

<h3>What is the engine piston displacement in liters of an engine whose displacement is listed as 430 in.²?</h3>

Generally, the equation for the dimensional analysis method,\, we convert in to L is mathematically given as

l*(v/l)*l/v

Therefore, piston displacement

PD=(450 in3) . (1 dm3 / 61.024 in3) . (1 L / 1 dm3)

PD= 7.37 L

In conclusion, piston displacement

PD= 7.37 L