C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.
<u>Explanation:</u>
As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.
So it is represented as,
R∝ l /A [ area is radius square]
So k is the proportionality constant used.
R = kl/A
Conductance is the inverse of resistance, so it is given as,
C= 1/R.
R₁ = kl₁ / A₁
R₂ = kl₂/A₂
R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]
= l₂A₁ / l₁A₂
If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12
So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.
Answer:
1.95g of Mg(OH)2 are needed
Explanation:
Mg(OH)2 reacts with HCl as follows:
Mg(OH)2 + 2 HCl → MgCl2 + 2H2O
<em>Where 1 mole of Mg(OH)2 reacts with 2 moles of HCl</em>
To solve this question we must find the moles of acid. Then, with the chemical equation we can find the moles of Mg(OH)2 and its mass:
<em>Moles HCl:</em>
158mL = 0.158L * (0.106mol / L) = 0.01675 moles HCl
<em>Moles Mg(OH)2:</em>
0.01675 moles HCl * (2mol Mg(OH)2 / 1mol HCl) = 0.3350 moles Mg(OH)2
<em>Mass Mg(OH)2 -Molar mass: 58.3197g/mol-</em>
0.3350 moles Mg(OH)2 * (58.3197g / mol) =
<h3>1.95g of Mg(OH)2 are needed</h3>
Answer:
Explanation:
Hello there!
In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:
It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:
Best regards!
The correct number of significant figures and digits is 3
Answer:
It’s either A or D buttttt probably d
Explanation:
