Answer:
840.18
Explanation:
Use the equation: Q = mcΔT
m = mass (5 g)
c = specific heat (4.18)
ΔT = change in temperature (65.3-25.1 = 40.2)
= 840.12
Answer:
D. For the atoms lower in the periodic table, the valence electrons are in higher energy levels and farther from the nucleus.
Explanation:
Atomic radius increases down the group because down the group, there an increase in the number of principle energy levels occupied. Now, these higher principal energy levels are made up of orbitals that are larger than the orbitals from the lower energy levels in size.
Therefore, the effect of this is that the greater number of principal energy levels will outweigh the increase in nuclear charge since nuclear charge also increases down the group and this in turn makes the atomic radius to increase as we go down the group.
Answer: C. For the atoms lower in the periodic table, the balance electrons are in higher energy levels and farther from the nucleus.
Explanation: As the valance electrons orbit farther from the nucleus the energy level increases from the top to the bottom of the periodic table. So the atoms lower in the periodic table, the balance electrons are in higher energy levels and farther from the nucleus, which result in an increase in the atomic radius.
Answer:
53.1 mL NaOH
Explanation:
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Explanation:
...................
O 3.04 x 1028
O 5.76 x 1029
O 6.02 x 1023
O 3.13
There are 5.78 × 10^20 sulfide ions in sodium sulfide
The formula of the compound is Na2S. The molar mass of the compound is calculated as follows; 2(23) +32 = 46 + 32 = 78 g/mol
Number of moles of Na2S = 0.15 g/ 78 g/mol = 0.0019 moles
Since there is 1 mole of Na^+ and 2 moles of S^2- in Na2S, the number of S^2- ions in 0.19 moles of Na2S is 0.00096 moles of S^2-.
If 1 mole of S^2- contains 6.02 × 10^23
0.095 moles of S^2- contains 0.00096 moles × 6.02 × 10^23/ 1 mole
= 5.78 × 10^20 sulfide ions
Learn more: brainly.com/question/1309057
Determine the amount of CO2(g) formed in the reaction if 8.00 grams of O2(g) reacts with an excess of C2H6(g) and the percent yield of CO2(g) is 90.0%.
Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
Given mass of oxygen gas = 8 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
For the given chemical equation:
By Stoichiometry of the reaction:
7 moles of oxygen gas produces 4 moles of carbon dioxide
So, 0.25 moles of oxygen gas will produce = of carbon dioxide
Now, calculating the mass of carbon dioxide from equation 1, we get:
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 0.143 moles
Putting values in equation 1, we get:
To calculate the experimental yield of carbon dioxide, we use the equation:
Percentage yield of carbon dioxide = 90 %
Theoretical yield of carbon dioxide = 6.292 g
Putting values in above equation, we get:
Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams
Answer:
C
Explanation: