<u>Option 1 </u>is a suitable graph for a “constant positive velocity”.
Option: 1
<u>Explanation</u>:
Consider a car is moving with the constant velocity, the position and time graph with “constant positive velocity” of the car is represented by the graph which is in option 1. In this graph (option: 1) has the straight line with the slope. The slope in this “position - time” graph represents the “speed of the car”, the velocity is positive hence the graph is drawn positive direction. Here, the “slope of the car” less so the car moves with less speed and constant positive velocity.
The correct answer is D. Using the law of conservation of mass the number of atoms on each side of the equation should be equal. Through introspection, we find that there are 2 atoms on reactant side as opposed 3Cl atoms on product side. If we add a coefficient of 3 on we get
.
Now there are 6Cl atoms on reactant side and 2 on product side, hence we add a coefficient of 2 on both and . The balanced chemical equation is,
Answer:
n ≈ 2.42 moles
General Formulas and Concepts:
<u>Chem</u>
Ideal Gas Law: PV = nRT
- P is pressure in mmHg
- V is volume in liters
- n is number of moles
- R is a constant (62.4 L · mmHg/mol · k)
- T is temperature in Kelvins
K = °C + 273
Explanation:
<u>Step 1: Define</u>
V = 50.2 L
P = 755 mmHg
T = -22.0°C = 251 K
<u>Step 2: Find moles </u><em><u>n</u></em>
(755 mmHg)(50.2 L) = n(62.4 L · mmHg/mol · k)(251 K)
37901 mmHg · L = n(15662.4 L · mmHg/mol)
n = 2.41987 moles
<u>Step 3: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules.</em>
2.41987 moles ≈ 2.42 moles
Answer:
Explanation:
Percent composition of sodium chloride .
NaCl
Mol weight = 23 + 35.5 = 58.5 g
58.5 g of NaCl contains 23 g of Na
100 g of NaCl contains 23 x 100 / 58.5 g of Na
= 39.3 %
Na = 39.3 %
58.5 g of NaCl contains 35.5 g of Cl
100 g of NaCl contains 35.5 x 100 / 58.5 g of Na
= 60.7 %
Cl = 60.7 % .
Answer:
Approximately 5.05 g of CO2 are used when 6.50 g of O2 are produced.
Explanation:
The answer to this question is the implication that the production of 6.5g of O2 requires 5g-6g CO2, since it is 63% oxygen by mass and 37% carbon dioxide by mass (not including water vapor). This would mean that 4g-5g CO2 productions could be expected to yield 2.25 or so grams oxygen for each gram CO2 produced, depending on how low the temperature was at which the reaction took place (a product like dry ice produces only cold enough to use 1/16th its volume in volume rather than 1/200th).