Does the density of a liquid depend on its volume? Write your answer as a CER.

Answers

Answer 1

Answer:

Answer: Yes

Explanation:

Density of a liquid depend on its volume. This is because Density is mass of liquid divided by volume.

Density is inversely proportional to volume.

As density increases, volume decreases and vice versa. The density for water is 1g/ milliliter but it changes with changes in temperature or there are impurities dissolved in it. Ice is less dense that liquid water and it's the major reason it's float because it's volume is inversely proportional to it's density.

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100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant
Be sure to answer all parts.Calculate the percent composition by mass (to 4 significant figures) of all the elements in calciumphosphate (Ca3(PO4)2), a major component of bone.% Ca%P% 0

A sample of solid calcium hdroxide, Ca(OH)2 is allowed to stand in water until a saturated solution is formed. A titration of 75.00mL of this solution with 5.00 x 10-2 M HCl 36.6 mL of the acid to reach the end pointCa(OH)2 + 2HCl ? CaCl + 2H2O
What is the molarity?

Answers

Answer: The concentration of $Ca(OH)_2$ is 0.0122 M.

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\nM_1=5.00* 10^(-2)M=0.05M\nV_1=36.6mL\nn_2=2\nM_2=?M\nV_2=75mL$

Putting values in above equation, we get:

$1* 0.05* 36.6=2* M_2* 75\n\nM_2=0.0122M$

Hence, the concentration of $Ca(OH)_2$ is 0.0122 M.

On another planet, the isotopes of titanium have the following natural abundances. a. Isotope 46Ti Abundance 70.900% Mass(amu) 45.95263
b. Isotope 48Ti Abundance 10.000% Mass(amu) 47.94795
c. Isotope 50Ti Abundance 19.100% Mass(amu) 49.94479
d. What is the average atomic mass of titanium on that planet?
e. I got 46.9 amu but it is wrong.

Answers

Answer:

Average atomic mass = 46.91466 amu

Explanation:

Step 1: Data given

Isotopes of titanium

46Ti = 70.900% ⇒ 45.95263 amu

48Ti = 10.000 % ⇒ 47.94795 amu

50Ti = 19.100 % ⇒ 49.94479 amu

Step 2: Calculate the average atomic mass of titanium

Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479

Average atomic mass = 46.91466 amu

When an electron moves up to higher energy levels, the atom Choose... a photon of light whereas the atom Choose... a photon of light when an electron drops to a lower energy level. The photons emitted from an atom appear as

Answers

Answer:

Explanation:

When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.

this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.

Which one of these could be in the unknown anion "X" in this acid: H3Xcarbonate
fluorate
nitrogen
nitrite

Could you explain how to find this? The process?

Answers

Answer:

the answer is nitrogen

Choose the option below that is a characteristic of ketones. a. They contain a carbonyl group that exhibits sp hybridization around the carbon atom.
b. They contain a carbonyl group with a nonpolar carbon-oxygen bond.
c. The functional group of this type of compound must always be on the end of a carbon chain.
d. The functional group of this type of compound must always be in the middle of a carbon chain.

Answers

Answer:

Option d.

Explanation:

Ketones contain a carbonyl groups as a functional group, which is a carbon bonded to oxygen with a double bond. In a ketone, the carbon is always bonded to two carbon atoms:

R-C(=O)-R'

The carbon in the carbonyl group has a hybridization sp2 (3 hybrid orbitals with an unhybridized p orbital), where two of the orbitals form sigma (σ) bonds with the other two carbons (R-C-R') and the other hybrid orbital form a sigma bond with the oxygen (C-O). The unhybridized p orbital on the carbon atom is used to form a pi (π) bond with the oxygen, thus forming the double bond (C=O).

The bond of a carbonyl group is polar, because of the difference of the electronegativity between the carbon atom and the oxygen atom.

Hence, from all of the above we can discard the option a, (the carbonyl groups exhibits sp2 hybridization), the option b (carbon-oxygen bond is a bond polar) and the option c (the group must always be in the middle of a carbon chain, the groups that are always in the end, are a aldehyde groups).

Therefore, the correct option is d, the functional group of this type of compound must always be in the middle of a carbon chain.

I hope it helps you!

Answer:

d. The functional group of this type of compound must always be in the middle of a carbon chain.

Explanation:

Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. Write the empirical chemical formula of this compound?(A) Ca2PO4
(B) Ca3PO6
(C) Ca4P2O4
(D) Ca3P2O8 (or Ca3(PO4)2)
(E) CaPO4

Answers

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

$Moles\ Ca = (38.7g)/(40.08g/mol) =0.966\ mol\n\nMoles\ P = (19.9g)/(30.97g/mol) =0.643\ mol\n\nMoles\ O = (41.2g)/(16.00g/mol) =2.58\ mol$

Step 2: Calculate the molar ratio

$C = (0.966)/(0.643) =1.50\n\nP = (0.643)/(0.643) = 1.00\n\nO = (2.58)/(0.643) =4.00$

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

Final answer:

The mass percentage composition of a compound can be used to determine its empirical formula. For a compound with 38.7% calcium (Ca), 19.9% phosphorus (P), and 41.2% oxygen (O), the empirical formula is Ca3(PO4)2.