the answer is fresh water
Answer:
D) 0.50 mole of Ne
Explanation:
Given data:
Number of molecules of nitrogen = 3.0×10²³ molecules
Which sample contain same number of molecules as nitrogen= ?
Solution:
A) 0.25 mole of O₂
1 mole = 6.022×10²³ molecules
0.25 mol × 6.022×10²³ molecules / 1 mol
1.51×10²³ molecules
B) 2.0 moles of He.
1 mole = 6.022×10²³ molecules
2.0 mol × 6.022×10²³ molecules / 1 mol
12.044×10²³ molecules
C) 1.0 moles of H₂
1 mole = 6.022×10²³ molecules
D) 0.50 mole of Ne
1 mole = 6.022×10²³ molecules
0.50 mol × 6.022×10²³ molecules / 1 mol
3.0×10²³ molecules
Answer : The mass of the water in two significant figures is, 
Explanation :
In this case the heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of iron metal = 
= specific heat of water = 
= mass of iron metal = 32.3 g
= mass of water = ?
= final temperature of mixture = 
= initial temperature of iron metal = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
Therefore, the mass of the water in two significant figures is,
Answer:
a) f = 3.02x10¹⁵ s⁻¹, and λ = 99.4 nm.
b) 99.4 nm
Explanation:
a) The energy of radiation is given by:
E = h*f
Where h is the Planck constant (6.626x10⁻³⁴ J.s), and f is the frequency. To have the highest frequency, the energy must be the highest too, because they're directly proportional. So we must use E = -E1 = 20x10⁻¹⁹ J
20x10⁻¹⁹ = 6.626x10⁻³⁴xf
f = 3.02x10¹⁵ s⁻¹
The wavelenght is the velocity of light (3.00x10⁸ m/s) divided by the frequency:
λ = 3.00x10⁸/3.02x10¹⁵
λ = 9.94x10⁻⁸ m = 99.4 nm
b) To have the shortest wavelength, it must be the highest energy and frequency, so it would be the same as the letter a) 99.4 nm.
Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant
3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.
From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.
AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
![\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BCl%24%5E%7B-%7D%24%5D%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.292%20mmol%7D%7D%7B%5Ctext%7B50.%20mL%7D%7D%20%3D%20%5Ctextbf%7B0.066%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20concentration%20of%20chloride%20ion%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.066%20mol%2FL%7D%7D%24%7D)
