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arlik [135]
3 years ago
15

Mrs. Sail's class wanted to observe a local pond. The class collected some pond water to analyze. When they placed a drop of the

water under a microscope, they saw many microorganisms in the water. The pond water looked clear when they collected it in the jar, so what could explain the presence of these organisms? A) They must have grown in the jar overnight. B) The microorganisms are too small to be seen with the naked eye. C) Somehow, the jar was contaminated on the way from the pond to the class. D) The microorganisms only inhabit a very small section of the lake, and the class got lucky.
Chemistry
2 answers:
olasank [31]3 years ago
8 0

B) the microorganisms are to small too see with the naked eye

LiRa [457]3 years ago
3 0

B. its more of a reliable answer because it's true. You cant see it with the naked eye if you had to put in under a microscope.



I might be wrong.

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Calculate the Empirical Formula for the following compound:<br> 0.300 mol of S and 0.900 mole of O.
ziro4ka [17]

Answer:

\boxed {\boxed {\sf SO_3}}

Explanation:

An empirical formula shows the smallest whole-number ratio of the atoms in a compound.

So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.

<h3>1. Divide </h3>

The first step is division. We divide the amount of moles for both elements by the <u>smallest amount of moles</u>.

There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300

  • Sulfur: 0.300/0.300= 1
  • Oxygen: 0.900/0.300= 3

<h3>2. Write Empirical Formula</h3>

The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.

  • S₁O₃

This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.

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\bold {The \  empirical \ formula \ for \ the \  compound \ is  \ SO_3}}

6 0
3 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
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