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1 year ago

What is observed when magnesium is reacted with hydrochloric acid? choose all that apply

Chemistry

1 answer:

Zanzabum1 year ago

3 0

It is observed that when magnesium reacts with hydrochloric acid, it produces visible bubbles of hydrogen gas.

A vigorous reaction will occur giving rise of heat as it is an exothermic reaction. If heat is applied then it should be more vigorous.

Reaction is as follows:

Mg(s) + HCl (aq) --> MgCl2 (aq) + H2(g)

Magnesium reacts easily with HCl to produce H2 gas and magnesium ions, Mg2+, and heat. The reaction is exothermic, so it heats up quickly.

Mg(s) + HCl (aq) --> MgCl2(aq) + H2(g)

The net ionic equation :

Mg(s) + 2H+ --> Mg2 + H2(g)

If water is removed from the solution then white crystals of Mgcl2 is obtained.

Or in simple words,

2Mg + 2HCl -> 2Mg+ + 2Cl- + H2(gas)

The magnesium is attacked by the hydrochloric acid resulting in the magnesium dissolving into the solution resulting in a solution of magnesium chloride in hydrochloric acid and the production of hydrogen gas.

Learn more about magnesium here : brainly.com/question/5759562

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The net charge on a sulfide ion (s2-) is -2. explain how this ion obtains its charge

Anestetic [448]

Sulfur is a group six element in period 3 with atomic number 16 and an electronic configuration of 2:8:6. Therefore, to attain a stable configuration it requires to gain two electrons forming an ion with a charge of -2. The negative charge is due to the gaining of electrons.

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3 years ago

Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively)

Reika [66]

Answer:

Average atomic mass = 79.9034 amu

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

<u>For first isotope: </u>

% = 50.69 %

Mass = 78.9183 amu

<u>For second isotope: </u>

% = 49.31 %

Mass = 80.9163 amu

Thus,

$Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu$

$Average\ atomic\ mass=40.0036+39.8998\ amu$

<u>Average atomic mass = 79.9034 amu</u>

4 0

3 years ago

could you please explain operations with scientific notation using this example. 5 × 10³ + 4.3 × 10⁴ i dont understand how to so

TEA [102]

Explanation:

The scientific notation:

$a\times10^k$

where

$1\leq a$ and k is integer.

We have the example:

$(5\times10^3)+(4.3\times10^4)$

You can write the numbers in a "normal" form:

$5\times10^3=5\times1000=5000\\\\4.3\times10^4=4.3\times10000=43000$

Make the sum:

$5000+43000=48000$

And next write it in the scientific notation:

$48000=4\underbrace{8000}_{\leftarrow4}=4.8000\times10000=4.8\times10^4$

<h3>Other method:</h3>

You can add numbers in scientific notation if the power of tens in both number is the same.

Therefore you must convert the first number:

$5\times10^3=0.5\times10\times10^3=0.5\times10^4$

Now, you can make the sum:

$(5\times10^3)+(4.3\times10^4)=0.5\times10^4+4.3\times10^4=(0.5+4.3)\times10^4=4.8\times10^4$

4 0

3 years ago

What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of table salt, NaCl?

tia_tia [17]

Answer:

3 mol/L

Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

50.0 mL/ 1000 mL = 0.05 L

Now, you have to convert grams to moles in order to solve for molarity (M).

1.) On the periodic table find the molecular weights of Na and Cl.

Na= 22.99 g/mol Cl= 35.45g/mol

2.) Add them together to have their combined molecular weights.

22.99 + 35.45= 58.44 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

10.0 g/ 58.44 g = 0.17111 mol

4.)You are now going to plug in your answers into the equation for Molarity.

M= 0.17111 mol / 0.05 L = 3.4222

5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 3.4222 to 3 mol/L

Sorry this explanation is long let me know if you need a better more written out sample.

3 0

3 years ago

If the enantiomeric excess of a mixture is 75%, what are the % compositions of the major and minor enantiomer?

Maru [420]

Let us say that R is the major enantiomer, while S is the minor enantiomer, therefore the formula for enantiomeric excess (ee) is:

ee = (R – S) * 100%

Let us further say that the fraction of R is x (R = x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:

75 = (x – (1 – x)) * 100

75 = 100 x – 100 + 100 x

200 x = 175

x = 0.875

Summary of answers:

R = major enantiomer = 0.875 or 87.5%

<span>S = minor enantiomer = (1 – 0.875) = 0.125 or 12.5%</span>

7 0

4 years ago