Answer:
Explanation:
We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.
The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.
- q= 47.1 J
- m= 14.0 g
- ΔT= 1.80 °C
Substitute these values into the formula.
Multiply the 2 numbers in parentheses on the right side of the equation.
We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).
The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.
The heat capacity of the liquid is approximately 1.87 J/g°C.
Answer:
Homogeneous solutions are solutions with uniform composition and properties throughout the solution.
Explanation:
For example a cup of coffee, perfume, cough syrup, a solution of salt or sugar in water, etc. Heterogeneous solutions are solutions with non-uniform composition and properties throughout the solution.
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Answer:
A D F
Explanation:
Its right but its not in order But its A D and F
Answer:
The number of neutron in the Aluminium Isotope is :
B. 14
Explanation:
Isotopes : These are the atoms which have same atomic number but have different mass number.
<u>This image shows the average atomic mass of Al element because it is in decimals</u>.
Atomic mass = 26.98154
(Note : mass number of single isotope can never be in decimals)
It is the average of mass of different isotopes of Al
Major Isotopes of 
are :
......atomic mass = 26
.......atomic mass = 27
mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)
mass of 
Now calculate the neutron in
Number of neutron = mass number - atomic number
= 27 - 13
Number of neutron = 14
(Atomic mass is same as mass number)
