Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Opposite pairs form ionic bonds, due to this the answer is D - Li and Br; They have unlike charges.
Hey there!
I can't be sure my answers are the exact words, but it should be something along the lines of...
The amount of water on Earth is constant, but the form and location of the water changes as it moves through the water cycle.
This means that Earth has always had the same amount of water within in, along with it being the same water the whole time. No new water was introduced to our planet. Some of our water is liquid, some is solid, and some is gas. Some is deep in the soil and some is high up in the atmosphere. Some is in rain and some is in snow.
I hope this helps!
Answer:
They are all alkali earth metals.
Explanation:
Their valence shell each has 2 electrons. Also, they are all shiny, silvery-white, somewhat reactive metals at standard temperature and pressure. They form alkaline solutions, hydroxides, when reacting with water and their oxides are found in the earth’s crust.
Answer:
The answer to your question is 7.64 atm
Explanation:
Data
Volume 1 = V1 = 2.1 l
moles = 4.65
Pressure 1 = P1 = 3.82 atm
Volume 2 = Volume 1/2
Pressure 2 = ?
Process
1.- Calculate the new volume
Volume 2 = 2.10/2
Volume 2 = 1.05 l
2.- Use Boyle's law to find the Pressure
P1V1 = P2V2
-Solve for P2
P2 = P1V1 / V2
-Substitution
P2 = (3.82)(2.1) / 1.05
-Simplification
P2 = 8.022/1.05
-Result
P2 = 7.64 atm
