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I am Lyosha [343]

3 years ago

13

A 3.93 g sample of a laboratory solution contains 1.25 g of acid. what is the concentration of the solution as a mass percentage

?

1 answer:

WITCHER [35]3 years ago

4 0

Divide 1.25/3.93 and then multiply by 100 to get the percent. This equals 31.81%

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Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati

worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4 + Cl-

Firstly, we calculate the number of moles of the ammonia as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

NH4+ H2O ⇄ NH3 H3O+

I 0.242 0 0

C -X +x +X

E 0.242-X X X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b = K w

, where K w- the self-ionization constant of water, equal to

10 ^-14 at room temperature

This means that you have

K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈ 0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0

3 years ago

Which element in period 3 has three times the electronegativity value compared to that of li in period 2

victus00 [196]

Answer:

Nitrogen

Explanation:

Elements in period two includes lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

According to periodic trends, the electro negativity values are expected to increase across the period up to fluorine. Hence, as we go right wards, we encounter elements with higher electronegative values.

While lithium has an electronegative value of 1 , the electronegative value of element nitrogen is thrrr times this which is equal to three

7 0

3 years ago

Read 2 more answers

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

Best regards!

6 0

2 years ago

Look up the active ingredient in baking soda. Write a molecular and net ionic equation when that active ingredient is mixed with

Volgvan

Answer:

The active ingredients in baking soda (NaHCO3) are

$Na^+$ and $HCO3^-$

when Baking soda reacts with Acetic acid

Molecular equation

NaHCO3(aq) + CH3COOH(aq) → Na(CH3COO)(aq) + CO2(g) +H2O(l)

Ionic equation

$Na^+ + HCO3^- + CH3COOH$ → $Na^+ + CH3COO^- +CO2 + H2O$

as $Na^+$ is present on both sides so it will cancel out and the net ionic equation will be

$HCO3^-(aq) + CH3COOH(aq)$ → $CH3COO^-(aq) + CO2(g) + H2O(l)$

6 0

3 years ago

What particle is J.J Thomson credited with discovering

Ilya [14]

He was credited with discovering the subatomic particle also known as the electron in 1897.

6 0

3 years ago