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vampirchik [111]

3 years ago

12

What are the uses of glycerin?

1 answer:

Ilia_Sergeevich [38]3 years ago

6 0

Explanation:

This medication is used as a moisturizer to treat or prevent dry, rough, scaly, itchy skin and minor skin irritations (e.g., diaper rash, skin burns from radiation therapy). Emollients are substances that soften and moisturize the skin and decrease itching and flaking.

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What the atmosphere mean

Alex_Xolod [135]

The envelope of gases surrounding the earth or another planet.

3 0

3 years ago

Which of the following is true of group 6A?

USPshnik [31]

Answer:

The answer to your question is: Includes sulfur and gain two electrons

Explanation:

Includes Chlorine This option is wrong, Chlorine belongs to group VII.

Includes Sulfur This option is true, Group VI includes Oxygen, Sulfur, Selenium, Tellurium.

Gain 2 electrons . This option is true, Elements in group VI have six valence electrons so they gain to electrons to become estable.

Tend to form +2 ions This option is wrong, this elements form -2 ions

Have 5 valence electrons This option is wrong, this elements have 6 valence electrons.

7 0

2 years ago

O que é transferência de Calor?

irinina [24]

Answer:

I speak English can someone translate

Explanation:

7 0

2 years ago

An aqueous solution has 124 g of a salt dissolved in 1000 g of solution. What is the concentration of the salt in units of %?

tester [92]

Answer: 12.4%

Explanation:

124/1000 * 100 = 12.4%

3 0

2 years ago

You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye

dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

$V$ = 10 mL solution

$V_{added}$ = 12.20 mL

$V_{total}$ = 22.20 mL

then,

M 0.103 M of NaOH,

$V_{rinsed}$ = experiment will not be affected

$V_{total \ base}$ = 10.38 mL

Now,

⇒ mol of NAOH = MV

= $0.103\times 10.38$

= $1.06914 \ m$

Whether Sulfuric acid, then

⇒ $H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O$

⇒ $mol \ of \ acid =\frac{1}{2}\times \ mol \ of \ base$

⇒ $1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid$

Before any dilution:

$V_{sample} = 10 \ mL$

⇒ $M \ acid = \frac{m \ mol}{V}$

$=\frac{ 0.53457 }{10}$

$=0.053457 \ M$ (Sulfuric acid)

6 0

3 years ago