Answer:
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Explanation:
<u>1. Balanced molecular equation</u>
<u>2. Mole ratio</u>
<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:
Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
Answer:
PART A
Explanation:
3. A cylinder of compressed gas has a pressure of 4.882 atm on one day. The next
day, the same cylinder of gas has a pressure of 4.690 atm, and its temperature is
8°C. What was the temperature on the previous day in °C? Ans: 20°C.
Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.
Explanation:
Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.
A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.
A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.
B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.
B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.
Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.
Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.
The answer is B. Acid turns blue litmus paper to red
Explanation:
I think there' will be a decrease in volume of the air.
