1. the earth spining on its axis.
a.92.96 million mi
b.<span>When the sun's rays strike Earth's surface near the equator, the incoming solar radiation is more direct (nearly perpendicular or closer to a 90˚ angle). Therefore, the solar radiation is concentrated over a smaller surface area, causing warmer temperatures.
c.</span><span>The moon's gravitational pull on the Earth is the main cause of the rise and fall of ocean tides. The moon's gravitational pull causes two bulges of water on the Earth's oceans—one where ocean waters face the moon and the pull is strongest and one where ocean waters face away from the moon and the pull is weakest.
d. </span><span>about 1,000 miles per hour (1,600kph)</span>
Answer:
<em><u>Question 1.</u></em><em> → </em>The amount remained is 15.625 g after 5 half-lives (30 hours).
<u><em>Question 2. </em></u><em>→ </em>mass % of water =41.4 %
<u><em>Question 3. </em></u><em>→ </em>C. 64 grams.
Explanation:
<em><u>Question 1. </u></em>
It is known that the decay of isotopes and radioactive material obeys first order kinetics.
Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
Thus, 30 hours represent (30/6) 5 half-lives.
500 g →(first 1/2 life) 250 g →(second 1/2 life) 125 g →(third 1/2 life) 62.5 g →(fourth 1/2 life) 31.25 g →(fifth 1/2 life) 15.625 g
So, The amount remained is 15.625 g after 5 half-lives (30 hours).
<u><em>Question 2.</em></u>
- The formula of Copper (II) Fluoride tetra hydrate is CuF₂.4H₂O.
- the molar mass of Copper (II) Fluoride tetra hydrate is the sum of the atomic masses o each element.
- molar mass of 1 mol of CuF₂.4H₂O = 64 + 2*(19) + 8* (1) + 4*(16) = 174 g/mol.
- molar mass of 4 mol of H₂O = 8*(1) + 4*(16) = 72 g/mol
- 1 mol of Copper (II) Fluoride tetra hydrate → 4 mol H₂O
mass % of water = (mass of 4 mol of H₂O) / (mass of 1 mol of CuF₂.4H₂O ) *100 = (72 /174) *100 = 41.4 %
<em><u>Question 3. </u></em>
<u>Mass percentage</u> is defined as the mass of a solute divided by the total mass of the solution, multiplied by 100
- mass percent = ( mass of solute / mass of solution) *100
∴ 16 = ( mass of solute / 400 g) *100
mass of solute = (16* 400)/ 100 = 64 g.
So the right choice is C. 64 grams.
Hola!
[ refer the attachment. ]
Compare this diagram with that of yours.. :)
hope it helps!
Answer: Tfinal = 7.1°C
Explanation:
heat released or absorbed = mass × specific heat capacity × change in temperature
q = m × cg × ΔT (eqn 1)
Note: ΔT = (Tfinal - Tinitial)
(q = ? ΔHsoln=25.7kJ/mole = 25700J/mole; mass of solution, m = 100 + 35g = 135g; cg = specific heat capacity of water = 4.18 J°C-1g-1; ΔT = ? masss of solute, NH4NO3 = 35g, molar mass of solute, NH4NO3 = 80g)
molar enthalpy of solution, ΔHsoln
<em> = heat absorbed or released ÷ moles of solute, </em><em>n</em>
ΔHsoln = q ÷ n
q = ΔHsoln × n
<em>moles solute</em>, n = mass solute (g) ÷ molar mass solute (g mol-1)
moles of solute, n = 35g/80g/mol = 0.4375 moles
q = 25700J/mol × 0.4375 mol = 11243.75J
From equation 1 above, ΔT = q / (m × cg) = 11243.75J / (135 × 4.18 J°C-1g-1) = 19.9°C
Since the reaction is endothermic, Tinitial > Tfinal, therefore, Tfinal = Tinitial - ΔT = 27 - 19.9 = 7.1°C