Its just called a test tube holder
Answer:
D) 4.8 × 10² M/s
Explanation:
Let's consider the following generic reaction.
A → B
The rate law is:
v = k . [A]ⁿ
where,
v: velocity
k: rate constant
[A]: concentration of the reactant A
n: reaction order for A
<em>What is the velocity of a first-order reaction when the reactant concentration is 6 × 10⁻²M and the rate constant is 8 × 10³s⁻¹?</em>
v = k . [A]¹ = 8 x 10³s⁻¹ . 6 x 10⁻²M = 4.8 × 10² M/s
Answer:
n = 5
Explanation:
methane => 1 Carbon => CH₃
ethane => 2 Carbons => C₂H₆
propane => 3 Carbons => C₃H₈
butane => 4 Carbons => C₄H₁₀
pentane => 5 Carbons => C₅H₁₂
hexane => 6 Carbons => C₆H₁₄
heptane => 7 Carbons => C₇H₁₆ => CH₃(CH₂)₅CH₃
Octante => 8 Carbons => C₈H₁₈
Nonane => 9 Carbons => C₉H₂₀
Decane => 10 Carbons => C₁₀H₂₂
Answer:
1.01 atm
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₂
p₁V₁T₂ = p₂V₂ Divide each side by V₂
p₂ = p₁ × V₁/V₂ × T₂/T₁
Data:
p₁ = 1.05 atm; V₁ = 285 mL; T₁ = 15.8 °C
p₂ = ?; V₂ = 292 mL; T₂ = 11.2 °C
Calculations:
(a) Convert <em>temperatures to kelvins
</em>
T₁ = (15.8+273.15) K = 288.95 K
T₂ = (11.2+273.15) K = 284.35 K
(b) Calculate the <em>pressure
</em>
p₂ = 1.05 atm× (285/292) × (284.35/288.95)
= 1.05 atm × 0.9760 × 0.9840
= 1.01 atm
Answer:
<em>Nuclear Binding Energy (Bond Energy), Higher (Greater)</em>
Explanation:
<em>The energy required to overcome the attraction between two bonded nuclei and their shared electrons is called </em><em>Nuclear Binding</em><em> </em><em>Energy (Bond Energy)</em><em>. This quantity is a measure of bond strength since the stronger the bond, the </em><em>Higher (Greater) </em><em>the amount of energy required to break it.</em>
