All categories

3 years ago

A substance might freeze when thermal energy is ______ the substance.

Chemistry

2 answers:

inessss [21]3 years ago

7 0

Answer:

the answer is when the thermal energy is removed from the substance.

maxonik [38]3 years ago

6 0

I think it’s “Removed from”

You might be interested in

If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2

Vlad [161]

Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

1 mole of O2 occupy 22.4L at stp.

Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.

Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:

Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

3 0

3 years ago

15.50 g of NH4Cl reacts with an excess of AgNO3. In the reaction 35.50 g AgCl is produced. what is the theoretical yield of AgCl

Readme [11.4K]

Answer:

The answer to your question is 41.6 g of AgCl

Explanation:

Data

mass of NH₄Cl = 15.5 g

mass of AgNO₃ = excess

mass of AgCl = 35.5 g

theoretical yield = ?

Process

1.- Write the balanced chemical reaction.

NH₄Cl + AgNO₃ ⇒ AgCl + NH₄NO₃

2.- Calculate the molar mass of NH₄Cl and AgCl

NH₄Cl = 14 + 4 + 35.5 = 53.5 g

AgCl = 108 + 35.5 = 143.5 g

3.- Calculate the theoretical yield

53.5 g of NH₄Cl -------------------- 143.5 g of AgCl

15.5 g of NH₄Cl ------------------- x

x = (15.5 x 143.5) / 53.5

x = 2224.25 / 53.5

x = 41.6 g of AgCl

8 0

3 years ago

The diagram shows a simple lipid.

kifflom [539]

This is not a phospholipid as it does not contain a phosphate group at the end of the chain, and is not a triglyceride as there is no glyceryl moiety. Each carbon bonded to hydrogens makes the maximum number of C-H bonds possible, therefore there are no multiple bonds between carbons and the lipid is saturated. Therefore the answer is A.
Hope this helps!

7 0

3 years ago

Read 2 more answers

Which of the following is an incorrect name for an acid? A. acetic acid B. hydrocarbonate acid C. hydrocyanic acid D. sulfurous

just olya [345]

Answer:

The answer to your question is letter B

Explanation:

Incorrect name

A. acetic acid This name is correct for the acid with formula CH₃COOH

B. hydrocarbonate acid This is not the name for acid but for a molecule that has hydrogen and a metal.

C. hydrocyanic acid This name is correct for the inorganic molecule with formula HCN

D. sulfurous acid This name is correct and is the name of the inorganic molecule with formula H₂SO₃.

E. phosphoric acid This name is correct for the acid with formula H₃PO₄.

5 0

3 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0

3 years ago