Which is it that causes thunderstorms the stratosphere or the troposphere.

Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.

 If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.

 The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .

 the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.

 Wind shear is the change in wind direction or speed with height in an atmosphere.

Explanation:

Stratosphere

Explanation:

Small storms would be residing in the troposphere but stronger thunderstorms and hurricanes reside in the stratosphere.

hloride, what is the mass of calcium in

explanation:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol