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Kamila [148]
3 years ago
6

The maps above show a front as it moves east across the United States. Which of these describes the weather change that can be e

xpected for Amarillo, Texas?
A. The cold front will bring cooler temperatures and snow because the cold front is producing large amounts of energy and snow moving southeast.



B. The warm front will bring warmer temperatures with hot air for several days because it pushes the cold air out of the way.



C. The warm front will bring warmer temperatures with light rain for a few days because the warm air rises and cools producing rain.



D. The cold front will bring cooler temperatures with thunderstorms and rain, which causes the moisture in the warm air to condense and release large amounts of energy.

Chemistry
1 answer:
kobusy [5.1K]3 years ago
4 0
It’s D,the cold will bring cooler temperatures.....:)
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0.785 moles of N2,fill a balloon at 1.5 atm and 301 K.<br> What is the volume of the balloon?
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Answer:

V = 12.93 L

Explanation:

Given data:

Number of moles = 0.785 mol

Pressure of balloon = 1.5 atm

Temperature = 301 K

Volume of balloon = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

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V = volume

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R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will put the values.

V = nRT/P

V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm

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3 years ago
Balance the following chemical equation, then answer the following question.
julsineya [31]
Molar mass:

O2 = 31.99 g/mol
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<span>2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

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10.0 x 25 x 31.99 / 2 x 144.22 =

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hope this helps!


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3 0
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The equilibrium constant k is actually the ratio of the concentration of the products over the concentration of reactants at equilibrium. So if the concentration of products < concentration of reactants, therefore the constant k will be small. But if the concentration of products > concentration of reactants, the constant k will be large. In this case the value is too small (x10^-19), therefore we can say that the reaction favors the reactant side:

 

the equilibrium lies far to the left

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