<span>C2Br2
First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is
PV = nRT
where
P = pressure (1.10 atm = 111458 Pa)
V = volume (10.0 ml = 0.0000100 m^3)
n = number of moles
R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) )
T = Absolute temperature
Solving for n, we get
PV/(RT) = n
Now substituting our known values into the formula.
(111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol))
= (1.11458/2398.721652) mol
= 0.000464656 mol
Now let's calculate the empirical formula for this compound.
Atomic weight carbon = 12.0107
Atomic weight bromine = 79.904
Relative moles carbon = 13.068 / 12.0107 = 1.08802984
Relative moles bromine = 86.932 / 79.904 = 1.087955547
So the relative number of atoms of the two elements is
1.08802984 : 1.087955547
After dividing all numbers by the smallest, the ratio becomes
1.000068287 : 1
Which is close enough to 1:1 for me to consider the empirical formula to be CBr
Now calculate the molar mass of CBr
12.0107 + 79.904 = 91.9147
Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So
91.9147 g/mol * 0.000464656 mol = 0.042708701 g
0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is.
0.08541 / 0.0427087 = 1.99982673
1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>
The number of mole of Ca reacted is:
4.86 g Ca/ (40.08 g/mol Ca)= 0.121 mol Ca
Because Ca reacted completely with oxygen and there is 2 mol Ca, there is 1 mol O2 reacted.
Total mass of oxygen that reacted is:
0.121 mol Ca* (1mol O2/ 2 mol Ca)* (32 g O2/ 1 mol O2)= 1.94 g O2 reacted.
Hope this would help~
Answer : The volume of 3.0 M spinach solution added should be, 50 mL
Explanation :
Formula used :
where,
are the initial molarity and volume of spinach solution.
are the final molarity and volume of diluted spinach solution.
We are given:
Now put all the given values in above equation, we get:
Hence, the volume of 3.0 M spinach solution added should be, 50 mL
Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
Answer:
igneous rock CAN become sedimentary rock through a process called ROCK CYCLE.
Explanation:
Rocks can be defined as solid structures of minerals that are formed naturally over a period of time. They are grouped into three main types which includes the following:
- igneous rock
- sedimentary rocks and
- metamorphic rocks.
Rocks are capable of transforming from one type to another through a process known as rock cycle. There are two forces that brings about this process which includes:
- The internal force : this is the Earth’s internal heat engine, which moves material around in the core and the mantle and leads to slow but significant changes within the crust.
- The external force: this is the the hydrological cycle, which is the movement of water, ice, and air at the surface, and is powered by the sun.
Molten magma cools to form either extrusive igneous rock or intrusive igneous rock. With time they undergo weathering, eroded, transported, and then deposited as sediments which are being compressed and cemented into SEDIMENTARY ROCKS. Again through the above mentioned forces, different kinds of rocks are either uplifted, to be re-eroded, or buried deeper within the crust where they are heated up, squeezed, and changed into METAMORPHIC ROCK.
Therefore the material in this sedimentary rock found in Rhombus planet used to be in igneous rock deep in Rhombus's interior due to continuous rock cycling on the planet. I hope this helps, thanks.
