The Options are as follow,
<span> (1) CaCl</span>₂<span> (s) (3) CH</span>₃<span>OH (l)</span>
<span> (2) C</span>₂<span>H</span>₆<span> (g) (4) Cal</span>₂<span> (aq)</span>
Answer:
Option-1 is the correct answer.
Explanation:
As we know crystal formation is the property of solids. Therefore, in given options we are given with four different states of matter.
Option A, CaCl₂ is in a solid state , so it can exist in crystal form.
Option 2, C₂H₆ (Ethane) is in gas form, so it cannot form crystals.
Option 3, CH₃OH (Methanol) is present in liquid form, so it fails to form crystals.
Option 4, CaI₂, it is dissolved in water, Hence, it is in aqueous state, Therefore it also lacks crystal structure.
Answer:
(A) The work done by the system is -101.325J
(B) The workdone by the system is -90.75J
Explanation:
(A) Workdone = -PΔV
Given that A = 100cm2 = 0.01m2
distance d = 10cm = 0.1m
ΔV= Area × distance
ΔV= 0.01 ×0.1
ΔV = 0.001m3
P= external pressure = 1atm = 101325Pa
Workdone = -0.001 × 101325
W= - 101.325Pa m3
1Pam3 = 1J
Therefore W = - 101.325J
The work done on the system is -101.325J
(B) Workdone = -PΔV
Given that A = 50cm2 = 0.005m2
distance d = 15cm = 0.15m
ΔV= Area × distance
ΔV= 0.005×0.15
ΔV = 0.00075m3
P=121kPa = 121000Pa
W= - 121000 × 0.00075
W= -90.75Pa m3
1Pam3 = 1J
W = - 90.75J
The woekdone by the system is -90.75J
Answer:
The awnser is A.
Explanation:
I got it right on edgenuity. If im wrong sorry ;-;
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
Answer:
H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.
Explanation:
