The list shows the results of six athletes competing in the long jump, which has a target distance of 20 feet.

The element nickel has five naturally occurring isotopes. Which of the following describes the relationship of these isotopes?

Elden [556K]

<h3>Answer:</h3>

Different mass, same atomic number

<h3>Explanation:</h3>

Elements are made of similar atoms, however the atoms may have different number of neutrons but the same number of protons in their nucleus.
Such atoms that have equal number of protons but different number of neutrons are known as isotopes.
Difference in the number of neutrons makes isotopes to have different mass numbers.
Atomic number is the number of protons in the nucleus of an atom, therefore, isotopes have the same atomic number.

6 0

1 year ago

Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre

vaieri [72.5K]

Answer:

The pressure 10,000 m below the surface of the sea is 137.14 MPa.

The density 10,000 m below the surface of the sea is 2039 kg/m3

Explanation:

P0 and ρ0 are the pressure and density at the sea level (atmosferic condition). As the depth of the sea increases, both the pressure and the density increase.

We can relate presure and density as:

$\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\$

Rearranging

$\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\$

With this equation, we can calculate P at 10,000 m below the surface:

$P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa$

The density at 10,000 m below the surface of the sea is

$\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3$

4 0

1 year ago

Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T

GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0

1 year ago

The question is on the pic, thanks :)

Inessa05 [86]

It’s the BOA not the dog or kangaroo

8 0

1 year ago

) Starting with 5.00 g barium chloride n hydrate yields 4.26 g of anhydrous barium chloride after heating. Determine the integer

RSB [31]

Answer:

2

Explanation:

Mass of water molecule = mass of hydrated salt - mass of anhydrous salt

Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.

Number of moles = mass / molarmass

Molar mass of water = 18.015g/mol

No. of moles of water = 0.74 / 18.015 = 0.0411 moles.

Mass of BaCl2 present =?

1 mole of BaCl2 = 208.23 g

X mole of BaCl2 = 4.26 g

X = (4.26 * 1) / 208.23

X = 0.020

0.020 moles is present in 4.26g of BaCl2

Mole ratio between water and BaCl2 =

0.0411 / 0.020 = 2

Therefore 2 molecules of water is present the hydrated salt.

4 0

2 years ago