All categories

3 years ago

Explain how to use an acid-base indicator to determine pH

2 answers:

7 0

Acid-base indicator changes color based on pH.

drop some in a solution n watch the color changes. different indicators show different colors at different pH. they usually have standard colors for comparison.

Gala2k [10]3 years ago

5 0

Answer:

Litmus paper

Explanation:

for this very example your gonna need a litmus paper

You might be interested in

The image below represents which of the following (Check all that apply):

mihalych1998 [28]

What was he answer?? Please I am taking the test, what are they.

6 0

3 years ago

Read 2 more answers

The salt sodium sulfate, Na2SO4, can be formed by a reaction between...

ANTONII [103]

C. I asked my friend....

8 0

4 years ago

What is the mass, in grams, of 1.50 mol of iron (III) sulfate? Express your answer using three significant figures. mm = nothing

Paul [167]

Answer:

For A: The mass of iron (III) sulfate is 600. g

For B: The moles of ammonium carbonate is 0.07216 moles

For C: The mass of given number of molecules of aspirin is 0.359 grams.

For D: The molar mass of diazepam is 284.7 g/mol

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For A:

We are given:

Number of moles of iron (III) sulfate = 1.50 mol

Molar mass of iron (III) sulfate = 399.9 g/mol

Putting values in equation 1, we get:

$1.50mol=\frac{\text{Mass of iron (III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron (III) sulfate}=(1.50mol\times 399.9g/mol)=600.g$

Hence, the mass of iron (III) sulfate is 600. g

For B:

We are given:

Mass of ammonium carbonate = 6.935 g

Molar mass of ammonium carbonate = 96.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium carbonate}=\frac{6.935g}{96.1g/mol}=0.07216mol$

Hence, the moles of ammonium carbonate is 0.07216 moles

For C:

We are given:

Number of aspirin molecules = $1.20\times 10^{21}$

Mass of 1 mole of aspirin = 180.16 g/mol

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupies 1 mole

So, $6.022\times 10^{23}$ number of molecules of aspirin has a mass of 180.16 grams

Thus, $1.20\times 10^{21}$ number of molecules of aspirin will have a mass of $\frac{180.16g}{6.022\times 10^{23}}\times 1.20\times 10^{21}=0.359g$

Hence, the mass of given number of molecules of aspirin is 0.359 grams.

For D:

We are given:

Moles of diazepam = 0.05570 mol

Given mass of diazepam = 15.86 g

Putting values in equation 1, we get:

$0.05570mol=\frac{15.86g}{\text{Molar mass of diazepam}}\\\\\text{Molar mass of diazepam}=\frac{15.86g}{0.05570mol}=284.7g/mol$

Hence, the molar mass of diazepam is 284.7 g/mol

6 0

3 years ago

U What is the hydronium ion concentration of a substance with a pH of 2?

denpristay [2]

Answer:

The hydronium ion concentration of the substance is 0,01 M.

Explanation:

The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and greater than 7 to 14 basic. It is calculated as

pH = -log (H 30+), and in this case we solve to calculate (H30 +):

(H30 +) = antilog -pH= antilog -2

(H30+)=0,01 M

4 0

4 years ago

Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t

iragen [17]

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

E S S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

Solubility in 0.0120 M CoBr₂ (S')

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0.0240

C +S' +S'

E S' 0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

8 0

3 years ago