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vlada-n [284]
3 years ago
8

What is radiologist​

Chemistry
2 answers:
Ainat [17]3 years ago
6 0

Radiologists are medical doctors that treat injuries using medical imaging (radiology)

Dominik [7]3 years ago
3 0

Answer:

a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.

Explanation:

You might be interested in
If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate
NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0
3 years ago
How many moles of O2 are produced in the decomposition of 24 g of water in the
Natali5045456 [20]
M of water = 2(1) + 16 = 18 g/mol
n of water = 24g/(18g/mol) = 4/3 mol
n of H2O : n of O2 = 2 : 1
2:1 = 4/3 : x
x = (4/3)/2 = 2/3 mol of O2
8 0
3 years ago
PLEASE HELP ME ASAP!!!! WILL GIVE BRAINLIEST TO FIRST CORRECT ANSWER!!!!!!! THANKS:)
VikaD [51]

Answer:

Explanation:

1. How many molecules are present in 3 moles of silicon dioxide (sio2)

A. 5.00 x 10^23

B. 6.02 x 10^23

C. 12.36 x 10^23

D. 18.06 x 10^23

Answer:

18.066  × 10²³atoms

Explanation:

Given data:

Moles of silicon dioxide = 3 mol

Number of molecules = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

one mole = 6.022 × 10²³  atoms

3 mole × 6.022 × 10²³ atoms / 1mol

18.066  × 10²³atoms

2. What is the molar mass of Ag2S?

Answer:

Molar mass of  Ag₂S = 247.8014 g/mol

Explanation:

The molar mass of a chemical compound is sum of atomic mass of all constituent atoms.

For Ag₂S

Atomic mass of Ag = 107.8682 amu

Atomic mass of S = 32.065 amu

There are two silver atoms that's why

107.8682 amu × 2= 215.7364

Molar mass of Ag₂S:

Molar mass of  Ag₂S =  215.7364 + 32.065

Molar mass of  Ag₂S = 247.8014 g/mol

3) In the balanced equation 4KO + 2CO2 = 2K2CO3 + CO3 Which two chemicals have a molar ratio of 3:4?

Answer:

The equation:

4KO₂ + 2CO₂  → 2K₂CO₃ + 3O₂

In this balanced equation oxygen and potassium superoxide have ration 3:4.

Explanation:

Chemical equation:

4KO + 2CO₂  → 2K₂CO₃ + CO₃

The given equation does not have any two chemical ration 3:4.

The equation:

4KO₂ + 2CO₂  → 2K₂CO₃ + 3O₂

In this balanced equation oxygen and potassium superoxide have ration 3:4.

Coefficient with reactant and product.

Reactant                             Product

KO₂ = 4                             K₂CO₃ = 2

CO₂  = 2                            O₂ = 3

So KO₂ and oxygen have molar ratio 3 : 4 in balanced chemical equation.

7 0
4 years ago
Tell how many atoms of each element are in one molecule of the substance of table salt
Tatiana [17]

Answer:

Explanation:

Then, multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.Aug 15, 2020

5 0
3 years ago
Consider the reaction below that has a Keq of 8.98 X 10-2. The initial concentration of A and B is 1.68 M. Calculate the equilib
DaniilM [7]

Answer:

[C] = 0.4248M

Explanation:

                   A     +     B      ⇄     2C

C(i)          1.68M      1.68M         0.00

ΔC              -x            -x             +2x

C(eq)      1.68-x       1.68-x          2x

Keq = [C]²/[A][B] = (2x)²/(1.68 - x)²= 8.98 x 10⁻²

Take SqrRt of both sides => √(2x)²/(1.68 - x)² = √8.98 x 10⁻²

=> 2x/1.68 - x = 0.2895

=> 2x = 0.2895(1.68 - x)

=> 2x = 0.4863 - 0.2895x

=> 2x + 0.2895x = 0.4863

=> 2.2895x = 0.4863

=> x = 0.4863/2.2895 = 0.2124

[C] = 2x = 2(0.2124)M = 0.4248M in 'C'

3 0
3 years ago
Read 2 more answers
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