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GuDViN [60]
3 years ago
5

You wish to make a 0.405 M hydrochloric acid solution from a stock solution of 3.00 M hydrochloric acid. How much concentrated a

cid must you add to obtain a total volume of 100 mL of the dilute solution?
Chemistry
1 answer:
Vesnalui [34]3 years ago
4 0

Answer:

Explanation:

We have to make 100 mL 0f .405M HCl from 3 M solution of HCl .

volume of 3M to be taken required . Let this volume be V litre .

V litre of 3M will contain 3 V moles of HCl .

moles contained by 100 mL of ,405 HCl = .405 x .1 = .0405 moles .

So 3V = .0405

V = .0405 / 3 litre

= .0405 x 1000 / 4 mL

= 10.125 mL

So we have to take 10.125 mL of 3M HCl and add water to them to make its volume 100 mL .

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tatyana61 [14]

Answer: (3) They have different masses and the same charges.

Explanation:

Atoms are mainly composed of three main types of particles (in fact there are more particles that are smaller and make up these main particles): electrons, protons and neutrons.

In the specific case of  <u>electrons</u> and <u>protons</u>, <u>both have the same charge, but electrons have the opposite sign</u>. This means a proton has an electric charge of  +1 and an electron has an electric charge of  −1.

On the other hand, <u>protons have more mass than electrons.</u> In fact, thetex]mass of an electron is about approximately \frac{1}{2000} the mass of a proton, this means these negative charged particles contribute almost nothing to the total mass of an atom (unlike protons, which together with neutrons, make up virtually all of the mass of an atom).

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3 years ago
Given 40 grams of magnesium. How many mole is this
Dahasolnce [82]

To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.

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\frac{40 g Mg}{1} x\frac{1 mole Mg}{24.305 g Mg}

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In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.

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The balanced chemical reaction would be:

FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g) 

We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:

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The main idea of an essay is usually portrayed in the thesis.
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