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mafiozo [28]
3 years ago
12

What is the balanced chemical equation for the reaction of potassium with sulfur?

Chemistry
1 answer:
Aneli [31]3 years ago
8 0

K + S = K2S

Potassium reacts with sulfur to produce potassium sulfide

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D) The average mass of all isotopes of the element
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38.2mm converted into cm
lina2011 [118]
38.mm converted into cm would be 3.82
5 0
3 years ago
Calcium hypochlorite (ca(ocl)2, mw = 142.983 g/mol) is often used as the source of the hypochlorite ion (ocl–, mw = 51.452 g/mol
stiv31 [10]
When the concentration is expressed in ppm, that means parts per million. It is also equivalent to mg/L. For this problem, we do stoichiometric calculations. We manipulate the units by cancelling like units if they appear in the numerator and denominator side until we come with the amount of solid Ca(OCl)2 needed. The solution is as follows:

40 mg/L * (1 L/1000 mL) * 50 mL * (1 g/1000 mg) * (1 mol OCl⁻/51.452 g) * (1 mol Ca(OCl)₂/ 2 mol OCl⁻) * (142.983 g Ca(OCl)₂/mol) * 0.95 = 2.64×10⁻3 g or 2.64 mg.

Therefore, you would need 2.64 mg of solid Ca(OCl)₂.
8 0
4 years ago
Suppose you have 300.0 mL of a 0.450 M sodium hydroxide solution.
icang [17]

Answer:

135 moles

Explanation:

300*0.45

= 135

PLS GIVE BRAINLIEST

7 0
3 years ago
Read 2 more answers
1. A mixture of NaOH and Na2CO3 solution required 20.50 mL of 0.5 M HCl using
netineya [11]

Answer:

i) for NaOH (or KOH) = 13.75 mL

ii) for Na_{2}CO_{3} = 13.50 mL.

Explanation:

Phenolphthalein is an indicator which shows change in color when the conditions are highly basic.

Methyl orange is an indicator which shows change in color in presence of highly acidic medium.

For titration of NaOH and HCl we can use phenolphthalein but for sodium carbonate (a weak base) with HCl we use mehtyl orange.

Now in case of mixture of given strong base and weak base the reading or end point obtained from phenolphthalein, shows the neutralization of NaOH only and half neutralization of sodium carbonate.

NaOH+HCl--->NaCl+H_{2}O

Na_{2}CO{3}+HCl--->NaHCO_{3}+H_{2}O

While the reading or end point of methyl orange shows the neutralization of both the base present.

a) The volume of HCl used for phenolphthalein end point = 20.50 mL

Let us say the volume of HCl used for NaOH = V1

The volume of HCl used for half neutralization of sodium carbonate = V2

V1+V2 = 20.50.........(1)

b) the volume of HCl used for methyl organe end point = 27.25

This volume of HCl used for both NaOH and Na₂CO₃

V1+2V2 = 27.25  ... (2)

Equating equation 1 and 2

20.50-V2=27.25-2V2\\V2=6.75

i) Thus the volume of Acid used for NaOH (Or KOH if present in place of NaOH) = 20.50-6.75= 13.75 mL

ii) the volume of acid used for sodium carbonate =2X6.75= 13.5mL

4 0
3 years ago
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