10.0 g of calcium carbonate is treated with 10.0 g of HCl and,

if a solid dissolves into a liquid, what does it mean about the properties of its molecules? Describe what it would look like on

expeople1 [14]

Search it on the internet it will tell u

4 0

3 years ago

What is the total number of atoms in the following formula?<br>CaBr2

Furkat [3]

Answer:

Three

Explanation:

Ca= 1 atom

And Br2 stands for 2 Br atoms

6 0

3 years ago

Read 2 more answers

How many atoms are in each elemental sample?

Gnesinka [82]

Answer:

A. 2.9*10^24

B. 3.4*10^21

C. 1.2*10^25

D. 1.7*10^23

Explanation:

1 mol of any particle has 6.02*10^23.

A. 4.8 mol Cu* 6.02*10^23 (1/mol) ≈2.9*10^24

B. 5.6x10^-3 mol C *6.02*10^23 (1/mol) ≈3.4*10^21

C. 20.0 mol Hg*6.02*10^23 (1/mol)≈1.2*10^25

D. 0.285 mol Na*6.02*10^23 (1/mol)≈1.7*10^23

3 0

4 years ago

Evaporation of Cane Sugar Solutions. An evaporator is used to concentrate cane sugar solutions. A feed of 10 000 kg/d of a solut

Vera_Pavlovna [14]

Answer:

Weight of solution produced = 5135 kg

Amount of water removed = 4865 kg

Explanation:

For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg

mi = 0.38x10000 = 3800 kg

The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)

mo = 0.74xmt

0.74xmt = 3800

mt = 3800/0.74

mt = 5135 kg

This is the mass of solution produced.

The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :

wi = 10000 - 3800 = 6200 kg

wo = 5135 - 3800 = 1335 kg

wr = wi - wo

wr = 6200 - 1335

wr = 4865 kg

8 0

4 years ago

The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a

mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0

3 years ago