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Ivenika [448]
3 years ago
9

Which type of severe weather is NOT an intense tropical storm?

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer:  B.  Tornado

Explanation:  <u>A tornado is a violent rotating column of air extending from a thunderstorm to the ground. The most violent tornadoes are capable of tremendous destruction with wind speeds of up to 300 mph. They can destroy large buildings, uproot trees and hurl vehicles hundreds of yards. They can also drive straw into trees.</u>

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HeLP PLeaSe SEND HElP AAAAAAAAAAAAA
nevsk [136]

Answer:

Correct answer is Li

Explanation:

Lithium reacts with chlorine to form Lithium chloride.

$2Li+Cl_{2} \to 2LiCl$

This is a redox (Oxidation -Reduction) reaction

$2Li^{0} \to 2Li^{+2}+2e^{-}$ - Oxidation$Cl_{2}^{0}+2e^{-} \to 2Cl^{-}$  - Reduction

Hence the reducing agent in this reaction is Li

4 0
3 years ago
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Tanya [424]
<h3>Answer:</h3>

<u>Easy</u><u>!</u><u>!</u>

2+<u>1</u><u>7</u>=19

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4 0
2 years ago
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Determine the heat needed to warm 25.3 g of copper from 22 degrees celsius to 39 degrees celsius.
Serggg [28]

Answer:

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

Explanation:

Q=mc\Delta T

Where:

Q = heat absorbed  or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

We have mass of copper = m = 25.3 g

Specific heat of copper = c = 0.385 J/g°C

ΔT  = 39°C - 22°C = 17°C

Heat absorbed by the copper :

Q=25.3 g\times 0.385 J/g^oC\times 17^oC=165.59 J

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

5 0
4 years ago
The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appear
djverab [1.8K]

Answer: The rate of appearance of Br_2 is 0.0700Ms^{-1}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of HBr = -\frac{1d[HBr]}{2dt}Rate in terms of appearance of [tex]H_2 = \frac{1d[H_2]}{dt}

Rate in terms of appearance of Br_2 = \frac{1d[Br_2]}{dt}

-\frac{1d[HBr]}{2dt}=\frac{d[H_2]}{dt}=\frac{d[Br_2]}{dt}

Given :

-\frac{1d[HBr]}{dt}=0.140Ms^{-1}

The rate of appearance of Br_2;

\frac{1d[Br_2]}{dt}=-\frac{1d[HBr]}{2dt}=\frac{1}{2}\times 0.140=0.0700Ms^{-1}

Thus rate of appearance of Br_2 is 0.0700Ms^{-1}

6 0
3 years ago
Which of the following is/are considered alcohols?
Paladinen [302]

Answer: THE ANSWER IS D

Explanation:

an alcohol is a hydrocarbon chain with an hydroxyl group (OH)

A, B, AND C ARE ALL ALCOHOLS, SO THE ANSWER IS D

6 0
3 years ago
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