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3 years ago

1. Why is severe weather associated with low pressure systems?

2 answers:

8 0

1) low pressure systems tend to result in unsettled weather and may present clouds high winds and precipitation. As low pressure intensifies storms or hurricane may be formed

KonstantinChe [14]3 years ago

6 0

1. because the air continues to cool and the water vapor may condense into precipitation, which results in unsettled weather

2. air is more dense in a region of high pressure, so the air pushes toward less dense regions. the air begins to warm as it descends, gets rid of the formation of clouds

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Which horizon would have the highest amount of organic material?

alexandr402 [8]

O horizon is the answer

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2 years ago

Can convection occur in both liquids and

Triss [41]

Answer:

Yes, Convection can occur in both liquids and gases

Explanation:

The Particle Theory suggests that Particles are always moving. Convection occurs in the breeze you feel, when the warm particles quickly move upwards resulting in cold air quickly sinking and creating a breeze. Convection also occurs when you boil water, the molecules of warm water quickly move to the top and cold water quickly moves to the bottom resulting in a circle the constantly keeps the water circulating, making it boil.

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3 years ago

Consider the following intermediate chemical equations.

Zinaida [17]

Answer:

The first choice, $-205.7\; \rm kJ$ .

Explanation:

Let the three reactions, where the enthalpy change were known, be called (1), (2), and (3).

The goal is to find the enthalpy change of the fourth equation. Assume that this equation can be written as $x \times (1) + y \times (2) + z \times (3)$ for some $x$ , $y$ , and $z$ (might not be whole numbers or take positive values.) Then, by Hess's Law, the enthalpy change of that reaction would be $x \cdot \Delta H_1 + y \cdot \Delta H_2 + z \cdot \Delta H_3$ .

To find these $x$ , $y$ , and $z$ , consider: what combination of reaction (1), (2), and (3) would give the fourth reaction?

Imagine that the coefficients are positive for all the reactants, and negative for all the products.

For example: in (1), $\rm H_2\; (g)$ has a coefficient of $2$ . However, since it is on the the product side of (1), its value should be $-2$ . Also, in (3)

Since there is no $\rm H_2\; (g)$ in the desired equation, the value of $x$ , $y$ , and $z$ should ensure that $-2x + z = 0$ .

Another example: $\rm CH_4\; (g)$ is on the reactant side of the first reaction. Its coefficient in the equation is $1$ , so that corresponds to $+1$ . Since $\rm CH_4$ is neither in (2) nor in (3), the value of

In the desired equation, $\rm CH_4\; (g)$ is on the reactant side with a coefficient of $1$ . As a result, the value of $x$ , $y$ , and $z$ should ensure that $x = 1$ .

One such equation can be found for each species in the reactions.

$\displaystyle \begin{cases}x = 1 & \left(\text{For $\mathrm{CH_4\; (g)}$}\right) \\ -x - y= 0 & \left(\text{For $\mathrm{C\; (s)}$}\right) \\ -2\, x + z = 0 & \left(\text{For $\mathrm{H_2\; (g)}$}\right) \\ y = -1 & \left(\text{For $\mathrm{CCl_4\; (g)}$}\right) \\ -2\, y + z= 4 & \left(\text{For $\mathrm{Cl_2\; (g)}$}\right) \\ -2\, z = -4 & \left(\text{For $\mathrm{HCl\; (g)}$}\right)\end{cases}$ .

Solve this system of equations for $x$ , $y$ , and $z$ (this approach works only if at least one solution exists.) In this case,

$\displaystyle \begin{cases}x = 1 \\ y = -1 \\ z = 2\end{cases}$ .

Calculate the enthalpy change of the desired reaction:

$\begin{aligned}\Delta H &= x\times \Delta H_1 + y \times \Delta H_2 + z \times \Delta H_3 \\ &= 1 \times 74.6 + (-1) \times 95.7 + 2 \times (-92.3) \\ &= -205.7\; \rm kJ\end{aligned}$ .