Answer:
Light fur mice can be easily spotted by predators among the dark rocks
Answer:
The initial temperature is 499 K
Explanation:
Step 1: Data given
initial volume = 12 cm3 = 12 mL
Final volume = 7 cm3 = 7mL
The final temperature = 18 °C = 291 K
Step 2: Calculate the initial temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 0.012 L
⇒with T1 = the initial volume = ?
⇒with V2 = the final volume 0.007 L
⇒with T2 = The final temperature = 291 K
0.012 / T1 = 0.007 / 291
0.012/T1 = 2.4055*10^-5
T1 = 0.012/2.4055*10^-5
T1 = 499 K
The initial temperature is 499 K
No matter how big or little a material is, its properties remain the same. Size, form, color, and mass are the qualities that distinguish a material from non-substances. All of these qualities, including size, shape, color, and mass, can be seen and measured. Some characteristics are physical, while others are chemical. Physical characteristics include mass, volume, density, and color. Viscosity and solubility are examples of chemical qualities.
Answer:
There is 17.1 kJ energy required
Explanation:
Step 1: Data given
Mass of ethanol = 322.0 grams
Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K
Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K
Specific heat capacity = 2.44 J/g*K
Step 2: Calculate energy
Q = m*c*ΔT
⇒ m = the mass of ethanol= 322 grams
⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K
⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K
Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ
There is 17.1 kJ energy required
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. 
Major Species at B:
1. 
2. 
Major Species at C:
1. 
Major Species at D:
1. 
2. 
Major Species at E:
1. 
Major Species at F:
1. 
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK
= 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H
]/[H2
]
pH = pK
= 10.33