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Agata [3.3K]
2 years ago
12

Why does a cold front cause intense weather

Chemistry
2 answers:
Katena32 [7]2 years ago
7 0
A cold front causes intense weather because winds will move towards each other along the front
DiKsa [7]2 years ago
5 0
There are several severe weather events that occur due to cold fronts. The reason being is because winds will move towards each other along the front. The angle of a cold front is also greater than that of the other types of fronts, which creates more lift in the atmosphere vertically
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Name the parts of an electromagnet
Oxana [17]

Answer:

The iron core, copper wire, and an electricity source.

Explanation: Me

7 0
3 years ago
What type of tectonic activity occurs in the San Andreas Fault?
kykrilka [37]

The answer is a strike-slip. More specifically a right-lateral strike-slip.

6 0
3 years ago
What is the volume of 40g of sugar given it’s listed density
lora16 [44]

Answer:

25.157 cm³

Explanation:

Data Given:

Mass of Sugar (m) = 40g

Density of sugar given in literature = 1.59 g/cm³

Volume of Sugar = ?

The formula will be used is

                              d = m/v ........................................... (1)

where

D is density

m is the mass

v is the volume

So

Rearrange the Equation (1)

                              d x v = m

                               v = m/ d         ................................................ (2)

put the given values in Equation  (2)

                       v = 40g / 1.59 g/cm³

                       v = 25.157 cm³

volume of 40 g of sugar = 25.157 cm³

8 0
3 years ago
How many pounds in a 3.00 l bottle of drinking water?
slava [35]

Answer:

             6.61 Pounds

Solution:

Step 1: Calculate Mass of Water as;

                        Density  =  Mass  ÷  Volume

Solving for Mass,

                        Mass  =  Density  ×  Volume   ------ (1)

As,

                        Density of Water  =  1 g.cm⁻³

And,

                        3 L of Water  =  3000 cm³

Putting values in equation 1,

                        Mass  =  1 g.cm⁻³  × 3000 cm³

                        Mass  =  3000 g

Step 2: Convert Grams into Pounds;

As,

                        1 Gram  =  0.002204 Pounds

So,

                        3000 Grams  =  X Pounds

Solving for X,

                      X =  (3000 Grams  ×  0.002204 Pounds)  ÷  1 Gram

                      X =  6.61 Pounds

6 0
3 years ago
rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
Sunny_sXe [5.5K]

Complete Question

The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

So when

      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

3 0
2 years ago
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