Answer:
A cold front forms when a cold air mass pushes into a warmer air mass. Cold fronts can produce dramatic changes in the weather. ... As the cold front passes, winds become gusty. There is a sudden drop in temperature, and also heavy rain, sometimes with hail, thunder, and lightning.
An occluded front forms when a warm air mass is caught between two cooler air masses. The warm air mass is cut ofl or occluded' from the ground. The occluded warm front may cause clouds and precipitation. A swirling center of low air pressure is called a cyclone.
Answer:
The correct answer is 930 grams platelets.
Explanation:
The volume of blood given is 1.89 pints. Total number of gallons in 1.89 pints is,
= 1.89/8 = 0.236 gallon
1 gallon comprise = 3.785 L
So, 0.236 gallon comprise = 0.236 * 3.785 L = 0.89 L
As mentioned that 1 Liter of blood comprise 1.04 kilograms of platelets. Therefore, 0.89 L of blood will contain = 1.04 * 0.89 = 0.93 Kg platelets
1 Kg contain 1000 grams. So, number of platelets in grams will be,
= 1000 * 0.93 = 930 grams platelets.
Answer:
0.1
Explanation:
We must first put down the equation of the reaction in order to guide our solution of the question.
2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)
Now from the question, the following were given;
Concentration of acid CA= ??????
Concentration of base CB= 0.299M
Volume of acid VA= 17.8ml
Volume of base VB= 24.7ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
SUBSTITUTING VALUES;
CA= 0.299 × 24.7 ×2 / 17.8×1
CA= 0.8298 M
But;
pH= -log[H^+]
[H^+] = 0.8298 M
pH= -log[0.8298 M]
pH= 0.1
The image of the bonds are missing, so i have attached it.
Answer:
A) - Sigma bond
-Sp³ and Sp³
- None
B) - Sigma and pi bond
- Sp² of C and p of O
- p of C and P of O
Explanation:
A) For compound 1;
- the molecular orbital type is sigma bond due to the end-to-end overlapping.
- Atomic orbitals in the sigma bond will be Sp³ and Sp³
- Atomic orbitals in the pi bond would be nil because there is no pi bond.
B) For compound 2;
- the molecular orbital type is sigma and pi bond
-Atomic orbitals in the sigma bond would be Sp² of C and p of O
- The Atomic orbitals in the pi bond will be; p of C and p of O
