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3 years ago

13

What does the prefix trans-indicate?

2 answers:

Tom [10]3 years ago

7 0

The carbons on either side of the double bond are pointed in opposite directions.

kogti [31]3 years ago

5 0

Answer: I think trans- prefix Definition of trans- (Entry 3 of 3) 1 : on or to the other side of : across : beyond transatlantic. 2a : beyond (a specified chemical element) in the periodic table transuranium

Explanation:

The prefix trans- and its variant tra-, which mean “across,” appear in many English vocabulary words, for example: transmit, transform, and trajectory. Consider the word translation, which is the carrying “across” from one language into another.

Hope it helps

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Inessa [10]

Answer:

Bones are the correct answer.

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2 years ago

In a simple distillation setup, what is the sequence of equipment from the bench top to the round bottom flask

Julli [10]

In a simple distillation setup, the sequence of equipment from the bench top to the round bottom flask is:

Thermometer
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Liebig condenser

Round bottom flask
Bunsen burner

<h3>What is Distillation?</h3>

This is the process in which a mixture is separated through selective boiling and condensation.

The distillation flask and liebig condenser are usually located above the round bottom flask in the set up.

Read more about Distillation here brainly.com/question/24553469

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2 years ago

Which atom has only one pair of electrons in its outermost energy level?

inysia [295]

Answer:

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3 years ago

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Dafna1 [17]

sorry i know this:

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2 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

4 years ago