Answer:
A = 0.7871g
B = 0.8417g
C = 0.3192g
D = 0.3897g
Explanation:
Hello,
The number of grams of sodium (Na) in 2g of NaCl is
Molar mass of NaCl = 58.44g/mol
Molar mass of Na = 23g
23g of Na = 58.44g of NaCl
x g of Na = 2g of NaCl
x = (2 × 23) / 58.44
x = 0.7871g
Therefore, 0.7871g of Na is present in 2g of NaCl
2.
2g of Na₃PO₄
Molar mass of Na₃PO₄ = 163.94g/mol
Molar mass of Na = 23g
(3 × 23)g of Na = 163.94 g of Na₃PO₄
x g of Na = 2g of Na₃PO₄
x = (2 × 69) / 163.94
x = 0.8417g
0.8417g of Na is present in 2g of Na₃PO₄
3.
2g of NaC₇H₅O₂
Molar mass of NaC₇H₅O₂ = 144.103g/mol
Molar mass of Na = 23g
23g of Na = 144.103g of NaC₇H₅O₂
x g of Na = 2g of NaC₇H₅O₂
x = (2 × 23) / 144.103
x = 0.3192g
0.3192g of Na is present in 2g of NaC₇H₅O₂
4. 2g of Na₂C₆H₆O₇
Molar mass of Na₂C₆H₆O₇ = 236.08g/mol
Molar mass of Na = 23g
(2 × 23)g of Na = 236.08g of Na₂C₆H₆O₇
x g = 2g of Na₂C₆H₆O₇
x = (46 × 2) / 236.08
x = 0.3897g
0.3897g is present in 2g of Na₂C₆H₆O₇
Answer:

Explanation:
Hello!
In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

Best regards!
Answer:
Presence of water molecules
Explanation:
Temperature decreases with an increase in altitude. At sufficiently high altitudes, the air is much colder than the air on the surface of the Earth. Water vapor has a low density and is capable to rise until it gets into a medium with a significantly low temperature and pressure to condense and produce water droplets out of water vapor. Those droplets then form clouds.
819 ml is the volume the gas will occupy at 30.0°C.
Explanation:
Data given:
initial volume of the gas V1 = 590 ml
initial temperature of the gas T1 = -55 degrees OR 218.15
final volume of the gas V2 = ?
final temperature of the gas T2 = 30 degrees OR 303.15
Charles' Law equation is used to calculate the volume of gas at 30 degrees from the data given in the question.
= 
V2 = 
V2 = 
V2 = 819 ml
The final volume of the gas would be 819 ml.
Answer:
I think it should be Carbon.