Answer:
The correct matching of the air mass and the letters in the word bank are given as follows;
1. Warm and humid ↔ D
2. Extremely cold and dry ↔ B
3. Cold and dry ↔ A
4. Cold and humid ↔ C
5. Warm and dry ↔ E
Where;
A Represents continental polar
B Represents Artic
C Represents Maritime Polar
D Represents Maritime Tropical
E Represents Continental Tropical
Explanation:
A. A continental polar is one that can be described as a cold and dry climate as the region is located at a further away from the oceanic water bodies that add humidity to the climate
B. The regions of the Artic and the Antarctic have very limited amount of precipitation every year because the air is very cold as well as dry
C. A polar climate is a cold climate region, while a maritime climate is humid.
Therefore, the maritime polar climate combines both cold and humid conditions
D. A warm and humid region has high rainfall and humidity, as such the maritime climate which are humid and the tropical climate, which are warm, combine to give a warm and humid climate
E. The continental tropical climate can be described as warm and dry, compared to the continental water bodies, due to the location being distant from and therefore, the absence of high moisture containing wind that comes from the oceans.
The length of time required for half of the radioactive atoms in a sample to decay is its <span>half-life. The correct option among all the options that are given in the question is the first option or option "A". The other choices are incorrect and can be easily neglected. I hope that this is the answer that has come to your help.</span>
Answer:
1.47 atm
Explanation:
Step 1: Given data
- Initial volume (V₁): 32.4 L
- Initial pressure (P₁): 1 atm (standard pressure)
- Initial temperature (T₁): 273 K (standard temperature)
- Final volume (V₂): 28.4 L
- Final temperature (T₂): 352 K
Step 2: Calculate the final pressure of the gas
We can calculate the final pressure of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
P₂ = P₁ × V₁ × T₂ / T₁ × V₂
P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm
Increase in heat, molecules start to escape and it turns to vapor
Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.
