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8_murik_8 [283]
2 years ago
13

Galena, a mineral of lead, is a compound of the metal with sulfur. Analysis shows that a 2.34-g sample of galena contains 2.03 g

of lead. Calculate the:
(a) mass of sulfur in the sample;
Chemistry
1 answer:
dolphi86 [110]2 years ago
7 0

The mass of sulfur in the sample is 0.31g

To calculate mass of sulfur in sample: given that amount galena is 2.34g and mass of lead is 2.03 g as galena contain both lead and sulfur the mass of sulfur = mass of galena- mass of lead , that is, 2.34- 2.03=0.31g.The molecular mass is a algebraic sum of atomic mass of each element in the compound thus the mass of the compound can be obtained by adding atomic of each element. Lead(II) sulphide occurs naturally as the mineral galena, often known as lead glance (PbS). It is a significant source of silver and the most significant lead ore.

To learn more about mass:

brainly.com/question/19694949

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Phenolphthalein indicator was added, and the solution in the flask was titrated with 0.215M NaOH until the indicator just turned
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In titration, the moles of acid equal moles of base. You were given that 22.75ml of 0.215M NaOH is used, so calculate the number of moles of that base the experiment used in total. After that because you know mol base = mol acid, whatever amount of base you use must be the total amount of acid present in the solution. You were given the volume of the acid, and you have just found the total mols of acid. Using these two information, solve for the concentration. And one more thing, even though I'm pretty sure it won't affect your answer, you should always convert things to the proper units. Since the concentration we're talking about in this problem is molarity, which has the unit mol/L, you should always have all of your numbers in these units. It just make it simpler and will not confuse you
3 0
3 years ago
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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
A pharmaceutical company wants to test the efficiency of its new drug production techniques so they run 3 shifts of production f
spayn [35]

The percentage yield of the  new production technique is 82.8%

<h3>What is the percentage yield?</h3>

Production is the procedure by which finished products are obtained form the raw materials. The production process involves the passing of raw materials through a certain procedure that involves the use of certain machines and equipment to give us the required products.

We are told in the question that there are three shifts;

Shift 1 produces 4562 grams

Shift 2 produces 5783 grams

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Average production from the three shifts = 4562 grams + 5783 grams +  5247 grams/3 = 5197 grams

The theoretical average yield is =  7000 grams + 7000 grams + 7000 grams/3 = 7000 grams

Now the percentage yield = actual yield/ theoretical yield * 100/1

percentage yield = 5197 grams/7000 grams * 100/1

percentage yield = 82.8%

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3 0
2 years ago
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As you go a group it is easier lose lose because the electrons are farther away from the nucleus and there is less attraction from the positive charges.

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One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w
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Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

similarly,

weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

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hence, vapor density of the gas is 16.128g.

Explanation:

4 0
3 years ago
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