Answer:
2. Isotope, one of two or more species of atoms of a chemical element with the same atomic number and position in the periodic table and nearly identical chemical behavior but with different atomic masses and physical properties. Every chemical element has one or more isotopes.
3. Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).
Explanation: Try rewording the questions when looking it up. Hope this helps
Answer:
The answer is in the explanation
Explanation:
The atomic weight of an element is defined as the mass of 1 mole of atoms of the element. To find the atomic weight of the first 30 elements we must see our periodic table:
# Element Atomic Weight
1 Hydrogen 1.008
2 Helium 4.0026
3 Lithium 6.94
4 Beryllium 9.0122
5 Boron 10.81
6 Carbon 12.011
7 Nitrogen 14.007
8 Oxygen 15.999
9 Fluorine 18.998
10 Neon 20.180
11 Sodium 22.990
12 Magnesium 24.305
13 Aluminium 26.982
14 Silicon 28.085
15 Phosphorus 30.974
16 Sulfur 32.06
17 Chlorine 35.45
18 Argon 39.948
19 Potassium 39.098
20 Calcium 40.078
21 Scandium 44.956
22 Titanium 47.867
23 Vanadium 50.942
24 Chromium 51.996
25 Manganese 54.938
26 Iron 55.845
27 Cobalt 58.933
28 Nickel 58.693
29 Copper 63.546
30 Zinc 65.38
Answer:
V H2O = 170.270 mL
Explanation:
- QH2O ( heat gained) = Qcoffe ( heat ceded)
⇒ Q = m<em>C</em>ΔT
∴ m: mass (g)
∴ <em>C</em>:<em> </em>specific heat
assuming:
- δ H2O = δ Coffe = 1.00 g/mL
- <em>C</em> H2O = <em>C</em> coffe = 4.186 J/°C.g....from literature
⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)
∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe
⇒ Q = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J
⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)
⇒ (26371.8 J)/(154.882 J/g) = m H2O
⇒ m H2O = 170.270 g
⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL
Answer:
The correct approach is "12.25°C".
Explanation:
Given:
Mass of lead,
mc = 245 g
Initial temperature,
tc = 300°C
Mass of Aluminum,
ma = 150 g
Initial temperature,
ta = 12.0°C
Mass of water,
mw = 820 g
Initial temperature,
tw = 12.0°C
Now,
The heat received in equivalent to heat given by copper.
The quantity of heat = 
then,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
