Answer:
1 mole represents 6.023×1023 particles.
1 mole of iodine atom= 6.023×1023
Given 127.0g of iodine.
no. of iodine atom = 1 mole of iodine
1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg
Given 48g of Mg = 2×6.023×1023
no. of Mg = 2 moles of Mg
1 mole of chlorine atom= 6.023× 1023
no. of chlorine atom = 35.5g of chlorine atom
Given 71g of chlorine atom=2× 6.023× 1023
no. of chlorine atom = 6.023×1023
2 moles of chlorine atom.
Given that 4g of hydrogen atom.
will be equal to 4 × 6.023 × 1023
no. of atoms of hydrogen= 4 moles of hydrogen atom.
Answer:
The correct answer is option 3. Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present
Explanation:
In any laboratory experiment, all the apparatus needed to carry out a particular experiment must be provided. In this case, our apparatus will be crude oil with ocean water and oil spill eater which is the enzyme used.
We can then run a test reaction of crude oil with ocean water over time with Oil Spill Eater present.
Answer:
ΔS=0.148 KJ/K
Explanation:
Given that
Q = 100 KJ
T₁=200°C
T₁=200+273 = 437 K
T₂=5°C
T₂=5 + 273 = 278 K
Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.
So the total change in entropy given as
ΔS= - Q/T₁ + Q/T₂
ΔS= - 100/473 + 100/278 KJ/K
ΔS=0.148 KJ/K
Answer:
18.84 g of silver.
Explanation:
We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole of sulphur contains 6.02×10²³ atoms.
1 mole of sulphur = 32 g
Thus,
32 g of sulphur contains 6.02×10²³ atoms.
Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.
From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.
Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.
This is illustrated below:
1 mole of silver = 6.02×10²³ atoms.
1 mole of silver = 108 g
108 g of silver contains 6.02×10²³ atoms.
Therefore, Xg of silver will contain 1.05×10²³ atoms i.e
Xg of silver = (108 × 1.05×10²³)/6.02×10²³
Xg of silver = 18.84 g
Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur
<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
