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3 years ago

11

How are parentheses used in chemical formulas

1 answer:

Liula [17]3 years ago

6 0

They make the symbols individual

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Which of these time measurements is the smallest?

Anuta_ua [19.1K]

Answer:

In order to answer the question, we convert the measurements given to a common base unit. For this case, we use seconds.

A) 0.02 seconds

B) 0.02 teraseconds x (10^12 s / 1 terasecond) = 2x10^10 seconds

C) 2,500 milliseconds x (1 s / 1000 ms) = 2.5 seconds

D) 25,000 nanoseconds x (1 x 10^-9 s / 1 nanosecond) = 2.5 x 10^-5 seconds

So ,The correct answer is option " D "

7 0

3 years ago

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¿Que es bacteria y un virus ?<br>es para hoy por favor ayúdeme

swat32

i) Las bacterias son organismos microscópicos unicelulares que prosperan en diversos entornos. Estos organismos pueden vivir en el suelo, el océano y dentro del intestino humano.

ii) Un virus es un agente infeccioso submicroscópico que se replica solo dentro de las células vivas de un organismo. Los virus infectan todas las formas de vida, desde animales y plantas hasta microorganismos, incluidas bacterias y arqueas.

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3 years ago

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Consider the balanced chemical equation that follows. You are asked to determine how many moles of water you can produce from 4.

matrenka [14]

Explanation:

According to law of conservation of mass, mass of the reactants is equal to the mass of products in a chemical equation. As mass can neither be created nor it can be destroyed but it can be transformed from one form to another.

As it is given that hydrogen and in excess oxygen is reacting that leads to the formation of water. Hence, the chemical reaction equation will be as follows.

$2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)$

Since, it is given that 4 mol of hydrogen is reacting with excess of oxygen and gives 2 moles of water.

Hence, number of moles of water produced is calculated as follows.

4 mol of $H_{2} \times \frac{2 mol of H_{2}O}{2 mol of H_{2}}$

= 4 moles of $H_{2}O$

Thus, we can conclude that 4 moles of water you can produce from 4.0 mol of hydrogen and excess oxygen.

3 0

3 years ago

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago

Convert eachh into scientific notation

mixer [17]

Answer:

0.00580 →

3000 →

0.000908 →

200. →

Explanation:

yay

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3 years ago

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