The image below represents which of the following (Check all that apply):

That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p

puteri [66]

Answer:

B 1.23 g/cc

Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.

Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.

A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.

B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.

C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.

D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.

5 0

2 years ago

A 10.63 g sample of mo2o3(s) is converted completely to another molybdenum oxide by adding oxygen. the new oxide has a mass of 1

Roman55 [17]

MoO2 First, lookup the atomic weights of the elements involved Atomic weight molybdenum = 95.94 Atomic weight oxygen = 15.999 Now calculate the molar mass of Mo2O3 2 * 95.94 + 3 * 15.999 = 239.877 g/mol Now determine how many moles of the original Mo2O3 you had 10.63 g / 239.877 g/mol = 0.044314378 mol Determine how much oxygen was added 11.340 g - 10.63 g = 0.71 g How many moles of oxygen was added 0.71 g / 15.999 g/mol = 0.044377774 mol Looking at the number of moles of oxygen added and the number of moles of the original compound, they're the same. So 1 oxygen atom was added to each molecule. Since the formula was Mo2O3, the new formula becomes Mo2O4. But since you're looking for the empirical formula, you need to reduce it. Both 2 and 4 are evenly divisible by 2, so the empirical formula becomes MoO2

7 0

2 years ago

Consider the element in the periodic table that is directly to the right of the element identified in part (a). Would the 1s pea

astraxan [27]

The question is incomplete. Here is the complete question.

The photoelectron spectroscopy is shwon below.

(a) Based on the photoelectron spectrum, identify the unknown element and write its electron configuration.

(b) Consider the element in the periodic table that is directly to the right of the element identified in part (a). Would the 1s peak of this element appear to the left of, the right of, or in the same position as the 1s peak of the element in part (a)? Explain your reasoning.

Answer and Explanation: Photoelectron Spectroscopy is a method of determinining the relative energy of electrons in atoms and molecules.

It is based on the photoelectric effect: when a radiation energy incides on a substance, an electron is ejected from it. If we know the kinetic energy of the ejected electron, known as photoelectrons, and the energy of the incident radiation, it is possible to find the energy of the electron in the substance.

The energy needed to eject an electron from the sample is called Binding Energy and in an atom, depends on which shell the electron is: valence eletrons (outermost shell), binding energy is lower; core eletrons (innermost shell), binding energy is highest.

In the graph, vertical axis shows 5 peaks for different energies. The peak closer to the origin, the leftmost peak, correspond to the 1s subshell, since their are closest to the nucleus, and so, has the highest binding energy.

Following from left to the right, we noticed:

First, second and fourth peaks has the same height;
Third peak's height is 3x higher than 1st, 2nd and 4th;
Fifth peak is one unit higher than first, second and fourth;

(a) Then, we can conclude the eletron configuration of the element is

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}$

which is Phosphorus with atomic number of 15.

(b) The element to the right of element P is Sulfur (S). The peak 1s of sulfur will appear in the same position as the 1s peak of Phosphorus, because the elements in the Periodic Table are grouped according to certain properties. Elements in the same horizontal line are elements in the same period, which one of the characteristics is they have the same total number of electron shells.

4 0

2 years ago

(45 pts) What is the theoretical yield (in g) of iron(III) carbonate that can be produced from 1.72 g of iron(III) nitrate and a

mote1985 [20]

Answer:

1.04g of iron III carbonate

Explanation:

First, we must put down the equation of reaction because it must guide our work.

2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)

From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.

Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate

Mass of iron III nitrate reacted= 1.72g

Molar mass of iron III nitrate= 241.88 g∙mol–1

Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles

From the equation of the reaction;

2 moles of iron III nitrate yields 1 mole of iron III carbonate

7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate

Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass

Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate

8 0

2 years ago

Read 2 more answers

Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80

Damm [24]

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)

= 1010 g

% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %

6 0

2 years ago