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AysviL [449]
4 years ago
8

A 250ml aqueous solution contains 45.1microgram of pesticide.express the pesticide concentration in:

Chemistry
1 answer:
olchik [2.2K]4 years ago
3 0

Answer:

1. 1.80x10⁻⁵ (w/w %).

2. 1.80x10⁻⁴ parts per thousand

3. 0.18 parts per million

Explanation:

The solution contains 45.1μg / 250mL.

1. Weight percent (100 times mass in grams of solute per gram of solution, as there are 250mL of water = 250g):

45.1x10⁻⁶g / 250g * 100 =

<h3>1.80x10⁻⁵ (w/w %)</h3>

2. Parts per thousand (mg of solute per g of solution).

45.1μg * (1x10⁻³mg / 1μg) = 0.0451mg.

0.0451mg / 250g =

<h3>1.80x10⁻⁴ parts per thousand</h3>

3. Parts per million (μg of solute per g of solution):

45.1μg / 250g =

<h3>0.18 parts per million</h3>
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How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO2.
Serhud [2]

Answer:

9.89 x 10²³ molecules H₂S

Explanation:

To find the molecules of H₂S, you need to (1) convert grams S to moles S (via the atomic mass of sulfur), then (2) convert moles S to moles H₂S (via the mole-to-mole ratio from equation coefficients), and then (3) convert moles H₂S to molecules H₂S (via Avogadro's Number). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value.

Atomic Mass (S): 32.065 g/mol

2 H₂S(s) + SO₂(g) -----> 3 S(s) + 2 H₂O(l)

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

79.0 g S           1 mole            2 moles H₂S          6.022 x 10²³ molecules
---------------  x  ---------------  x  ----------------------  x  -------------------------------------  =
                        32.065 g            3 moles S                          1 mole

=  9.89 x 10²³ molecules H₂S

4 0
2 years ago
A sample of nitrogen gas expands in volume from 16 L to 5.4 L at constant
xxTIMURxx [149]

a. W = 0 J

b. W = - 308.028 J

<h3>Further explanation</h3>

Given

Nitrogen gas expands in volume from 1.6 L to 5.4 L

Required

The work done

Solution

Isothermal :

W = -P . ΔV

Input the value :

a. At a vacuum, P = 0

So W = 0

b. At pressure = 0.8 atm

W = - 0.8 x ( 5.4 - 1.6)

W = -3.04 L.atm ( 1 L.atm = 101.325 J)

W = - 3.04 x 101.325

W = - 308.028 J

6 0
3 years ago
Convert 3.82 inches to km
weeeeeb [17]

the answer is 0.000097 KM

6 0
4 years ago
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liraira [26]

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3 years ago
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A sample of air was collected on a day when the total atmosphere pressure was
zhuklara [117]

Answer:

1. The gas law used: Dalton's law of partial pressure.

2. Pressure of nitrogen = 331 mmHg

Explanation:

From the question given above, the following data were obtained:

Total pressure (Pₜ) = 592 mmHg

Pressure of Oxygen (Pₒ) = 261 mmHg

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The pressure of nitrogen in the sample can be obtained by using the Dalton's law of partial pressure. This is illustrated below:

Pₜ = Pₒ + Pₙ

592 = 261 + Pₙ

Collect like terms

592 – 261 = Pₙ

331 = Pₙ

Pₙ = 331 mmHg

Therefore, the pressure of nitrogen in the sample is 331 mmHg

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