The answer you are looking for is A: Adding more batteries will increase the current
Since you have a circuit in series, that means that all of the bulbs would be connected to one wire. If you were add more batteries to that <em>one </em>wire, you would end up with a greater flow of energy throughout the bulbs/wire
Answer:
0.454moles
Explanation:
Calculate the molar mass of AgBr
Ag=108
Br=80
Molar mass= 108+80
Molar mass=188g/mol
It's mole to gram question
2moles of Na2S2O3= 188g of AgBr
X moles of Na2S2O3=42.7g of AgBr
Cross multiply
188x=2×42.7
X= 85.4/188
X=0.454moles
Answer:
4.81×10¹⁰ atoms.
Explanation:
We'll begin by converting 3.2 pg to Ca to grams (g). This can be obtained as follow:
1 pg = 1×10¯¹² g
Therefore,
3.2 pg = 3.2 pg × 1×10¯¹² g / 1 pg
3.2 pg = 3.2×10¯¹² g
Therefore, 3.2 pg is equivalent to 3.2×10¯¹² g
Next, we shall determine the number of mole in 3.2×10¯¹² g of Ca. This can be obtained as follow:
Mass of Ca = 3.2×10¯¹² g
Molar mass of Ca = 40.08 g/mol
Mole of ca=.?
Mole = mass /molar mass
Mole of Ca = 3.2×10¯¹² / 40.08
Mole of Ca = 7.98×10¯¹⁴ mole.
Finally, we shall determine the number of atoms present in 7.98×10¯¹⁴ mole of Ca. This can be obtained as illustrated below:
From Avogadro's hypothesis,
1 mole of Ca contains 6.02×10²³ atoms.
Therefore, 7.98×10¯¹⁴ mole of Ca will contain = 7.98×10¯¹⁴ × 6.02×10²³ = 4.81×10¹⁰ atoms.
Therefore, 3.2 pg of Ca contains 4.81×10¹⁰ atoms.
Answer:
The atomic radius corresponds to Sigma
Explanation:
It is the Van Der Waals radius ;)
