Answer: There are 105 grams of acetylene in the tank.
Explanation:
According to ideal gas equation:
P = pressure of gas = 1.39 atm
V = volume of gas = 70.0 L
n = number of moles = ?
R = gas constant =
T = temperature of gas in Kelvin =
Mass of ethylene , M = Moles × Molar mass = 4.04 mole × 26 g/mol =105g
Thus there are 105 grams of acetylene in the tank.
Answer:
The periodic table is organized into groups (columns) and periods (rows).
Explanation:
The periodic table is organized into groups (columns) and periods (rows). The first ones contain those elements that have the same valence and have similar characteristics to each other. Example: group IA have valence 1 (they require 1 electron to reach the octet). The exception to this are the noble gases located in the last group to the right of the table, which have already filled their last energy level.
Periods have elements that have the same number of orbitals and have different properties; although the masses are similar.
Answer:
The problem provides you with the solubility of potassium chloride,
KCl
, in water at
20
∘
C
, which is said to be equal to
34 g / 100 g H
2
O
.
This means that at
20
∘
C
, a saturated solution of potassium chloride will contain
34 g
of dissolved salt for every
100 g
of water.
As you know, a saturated solution is a solution that holds the maximum amount of dissolved salt. Adding more solid to a saturated solution will cause the solid to remain undissolved.
In your case, you can create a saturated solution of potassium chloride by dissolving
34 g
of salt in
100 g
of water at
20
∘
C
.
Now, your goal here is to figure out how much potassium chloride can be dissolved in
300 g
of water at this temperature. To do that, use the given solubility as a conversion factor to take you from grams of salt to grams of water
300
g H
2
O
⋅
34 g KCl
100
g H
2
O
=
102 g KCl
You should round this off to one sig fig, since that is how many sig figs you have for the mass of water
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
mass of KCl
=
100 g
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−
Explanation:
Production of materials and transportation are the examples of three carbon emission.
Extraction and production of purchased materials and transportation of purchased fuels are the examples of three carbon emission. Scope 3 emissions refers to all indirect emissions that occur in the chain of the reporting company that is included in both upstream and downstream emissions.
Big machineries are used for the production and extraction of materials as well as the transportation requires fossil fuels for working which releases carbondioxide gas in the atmosphere so we can conclude that production of materials and transportation are the examples of three carbon emission.
Answer:
C₇H₁₄O₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1.152 g
Mass of CO₂ = 2.726 g
Mass of H₂O = 1.116 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen present in the compound. This can be obtained as follow:
For carbon (C):
Mass of CO₂ = 2.726 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Mass of C = 12/44 × 2.726
Mass of C = 0.743 g
For hydrogen (H):
Mass of H₂O = 1.116 g
Molar mass of H₂O = (2×1) + 16
= 2 + 16
= 18 g/mol
Mass of H = 2/18 × 1.116
Mass of H = 0.124 g
For oxygen (O):
Mass of compound = 1.152 g
Mass of C = 0.743 g
Mass of H = 0.124 g
Mass of O =?
Mass of O = (Mass of compound) – (Mass of C) + (Mass of H)
Mass of O = 1.152 – (0.743 + 0.124)
Mass of O = 1.152 – 0.867
Mass of O = 0.285 g
Finally, we shall determine the empirical formula for the compound as follow:
C = 0.743 g
H = 0.124 g
O = 0.285 g
Divide by their molar mass
C = 0.743 / 12 = 0.062
H = 0.124 / 1 = 0.124
O = 0.285 / 16 = 0.018
Divide by the smallest
C = 0.062 / 0.018 = 3.44
H = 0.124 / 0.018 = 7
O = 0.018 / 0.018 = 1
Multiply by 2 to express in whole number.
C = 3.44 × 2 = 7
H = 7 × 2 = 14
O = 1 × 2 = 2
Empirical formula => C₇H₁₄O₂