Answer:
<em>Low tides would be lower and high tides would be higher and any low lying coastline would be flooded</em>
Explanation:
<em>Also...</em>
<em>If the moon got about 20 times closer it would make a gravitational force 400 times greater than what we are used to now</em>
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.
C + O2→ CO2
It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>
C + O2→ CO2
In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.
In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.
<u>No of moles = mass of the substance÷molar mass of the substance</u>
No of moles of carbon = 14.4 /12= 1.2 moles
No of moles of O2 = Mass of reacted O2/Molar mass of O2.
No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32
No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.
Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.
Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2
Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2
Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g
<u>Thus 52.8 g of CO2 is produced.</u>
LiBr.
<h3>Explanation</h3>
Note that the group number in this answer refers to the new IUPAC group number, which ranges from 1 to 18. Counts from the left. Start with the first two column (group 1 and 2), go on to the transition elements (Sc, Ti, etc. in group 3 through 12), and continue with the nonmetals (group 13 through 18).
Li is a group 1 metal. As a metal, it tends to form positive ions ("cations"). Metals in group 1 and 2 are <em>main group</em> metals. The charge on main group metal ions tends to be the same as the group number of the metal. Li is in group 1. The charge on an Li ion will be +1. Formula of the Li ion will be 
.
Br is a group 17 nonmetal. As a nonmetal, it tends to form negative ions ("anions"). The charge on nonmetal ions excepting for H tends to equal the group number of the nonmetal minus 18. Br is in group 17. The charge on a Br ion will be 17 - 18 = -1. Formula of the Br ion will be 
All the ions in an ionic compound carry charge. However, some of the ions like 
are positive. Others ions like are negative. Charge on the two types of ions balance each other. As a result, the compound is <em>overall</em> neutral.
1 × (+1) + 1 × (-1) = 0. The positive charge on one ion balances the negative charge on one ion. The two ions would pair up at a 1:1 ratio.
The empirical formula for an ionic compound shows all the ions in the compound. Positive ions are written in front of negative ions. is positive and is negative. The formula shall also show the simplest ratio between the ions. For the compound between Li and Br, a 1:1 ratio will be the simplest. The "1" subscript in an empirical formula can be omitted. Hence the formula: LiBr.
