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2 years ago

Why are numbers placed before the elements and molecules in achemical equation?

Chemistry

1 answer:

Neko [114]2 years ago

7 0

Answer:

the are coefficients for the atom. the tell how many there are.

Explanation:

they tell you how many atoms there are of its type.

ex. 4 C²H³. there are 4 of the c² and h³ (couldn't find a way to use subscript so used the power signs to show.)

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46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride

slega [8]

Answer:

$m_{HgCl_2}=42.7gHgCl_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4$

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

$m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2$

Best regards.

3 0

2 years ago

What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4

Ivenika [448]

Answer:

130ml of HCl(36%) in 4.90L solution => pH = 1.50

Explanation:

Need 4.90L of HCl(aq) solution with pH = 1.5.

Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺

[HCl(36%)] ≅ 12M in HCl

(M·V)concentrate = (M·V)diluted

12M·V(conc) = 0.032M·4.91L

=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.

Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.

5 0

3 years ago

A solution is made by adding 29.1 mL of concentrated perchloric acid ( 70.5 wt% , density 1.67 g/mL ) to some water in a volumet

Lera25 [3.4K]

Answer:

The concentration of the solution is 1.364 molar.

Explanation:

Volume of perchloric acid = 29.1 mL

Mass of the solution = m

Density of the solution = 1.67 g/mL

$m=1.67 g/mL\times 29.1 mL=48.597 g$

Percentage of perchloric acid in 48.597 solution :70.5 %

Mass of perchloric acid in 48.597 solution :

= $\frac{70.5}{100}\times 48.597 = 34.261 g$

Moles of perchloric acid = $\frac{34.261 g}{100.46 g/mol}=0.3410 mol$

In 29.1 mL of solution water is added and volume was changed to 250 mL.

So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)

$Molarity=\frac{Moles}{Volume (L)}$

$=\frac{0.3410 mol}{0.250 L}=1.364 M$

The concentration of the solution is 1.364 molar.

6 0

2 years ago

Sodium chloride is an example of a

Andrei [34K]

C. Weak acid is the correct answer

6 0

3 years ago

Read 2 more answers

If an ideal gas has a pressure of 5.49 atm, a temperature of 88.78 °c, and has a volume of 22.03 l, how many moles of gas are in

kirza4 [7]

N = (PV)/RT
(T = 88.78 + 273 = 361.78K)
(R = 22.4/273 = 0.082)
= (5.49 x 22.03)/(0.082 x 361.78) = ?
Put it into the calculator. It's hard to do that on a mobile phone.

7 0

3 years ago

Why are numbers placed before the elements and molecules in achemical equation?​

Why are numbers placed before the elements and molecules in achemical equation?