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:
MgO.
Explanation:
charges of both satisfy one another (balanced) -- producing a compound MgO.
Answer:24.31
Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97
Contribution made by isotope of mass 24.99=24.99×10.00=249.9
Contribution made by isotope of mass 25.98=25.98×11.01=286.04
Total contribution=1894.97+249.9+286.04=2430.91
Average mass=2430.91÷100
=24.31
Answer:
a. liquid
b. solid
c. gas, (should be at it's boiling point)
Explanation: If the normal melting point of a substance is below room temperature, the substance is a liquid at room temperature. Benzene melts at 6°C and boils at 80°C; it is a liquid at room temperature. If both the normal melting point and the normal boiling point are above room temperature, the substance is a solid.
if you need an explanation to each lmk
Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min
