Answer: The new volume at different given temperatures are as follows.
(a) 109.81 mL
(b) 768.65 mL
(c) 18052.38 mL
Explanation:
Given:
= 571 mL, ![T_{1} = 26^{o}C](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D%2026%5E%7Bo%7DC)
(a) ![T_{2} = 5^{o}C](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%205%5E%7Bo%7DC)
The new volume is calculated as follows.
![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C%5Cfrac%7B571%20mL%7D%7B26%5E%7Bo%7DC%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7B5%5E%7Bo%7DC%7D%5C%5CV_%7B2%7D%20%3D%20109.81%20mL)
(b) ![T_{2} = 95^{o}F](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%2095%5E%7Bo%7DF)
Convert degree Fahrenheit into degree Cesius as follows.
![(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C](https://tex.z-dn.net/?f=%281%5E%7Bo%7DF%20-%2032%29%20%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D%20%3D%20%5E%7Bo%7DC%5C%5C%2895%5E%7Bo%7DF%20-%2032%29%20%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D%20%3D%2035%5E%7Bo%7DC)
The new volume is calculated as follows.
![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C%5Cfrac%7B571%20mL%7D%7B26%5E%7Bo%7DC%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7B35%5E%7Bo%7DC%7D%5C%5CV_%7B2%7D%20%3D%20768.65%20mL)
(c) ![T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%201095%20K%20%3D%20%281095%20-%20273%29%5E%7Bo%7DC%20%3D%20822%5E%7Bo%7DC)
The new volume is calculated as follows.
![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C%5Cfrac%7B571%20mL%7D%7B26%5E%7Bo%7DC%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7B822%5E%7Bo%7DC%7D%5C%5CV_%7B2%7D%20%3D%2018052.38%20mL)
Answer : The
must be administered.
Solution :
As we are given that a vial containing radioactive selenium-75 has an activity of
.
As, 3.0 mCi radioactive selenium-75 present in 1 ml
So, 2.6 mCi radioactive selenium-75 present in ![\frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L](https://tex.z-dn.net/?f=%5Cfrac%7B2.6mCi%7D%7B3.0mCi%7D%5Ctimes%201ml%3D0.86666ml%5Ctimes%201000%3D866.66%5Cmu%20L)
Conversion :
![(1ml=1000\mu L)](https://tex.z-dn.net/?f=%281ml%3D1000%5Cmu%20L%29)
Therefore, the
must be administered.
Answer:
two north poles and two south poles
Explanation:
A single magnet has a north pole and a south pole. If it is broken into two pieces, then each of the two pieces will have a north pole and a south pole.
No matter how many times or into how many pieces a magnet is broken, the resulting pieces will have two poles each.
Answer:
The answer to your question is <u>111 g of CaCl₂</u>
Explanation:
Reaction
2HCl + CaCO₃ ⇒ CaCl₂ + CO₂ + H₂O
Process
1.- Calculate the molecular mass of Calcium carbonate and calcium chloride
CaCO₃ = (1 x 40) + (1 x 12) + ((16 x 3) = 100 g
CaCl₂ = (1 x 40) + (35.5 x 2) = 111 g
2.- Calculate the amount of calcium chloride produced using proportions.
The proportion CaCO₃ to CaCl₂ is 1 : 1.
100 g of CaCO₃ ------------- 111 g of CaCl₂
Then 111g of CaCl₂ will be produced.
Answer:
The minimum concentration of Cl⁻ that produces precipitation is 12.6M
Explanation:
The Ksp of PbCl₂ is expressed as:
PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)
The Ksp is:
Ksp = 1.6 = [Pb²⁺] [Cl⁻]²
When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.
A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:
1.6 = [0.010M] [Cl⁻]²
160 = [Cl⁻]²
12.6M = [Cl⁻]
<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>