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3 years ago

Can digital formats hold more data than analog

Chemistry

1 answer:

Oksi-84 [34.3K]3 years ago

3 0

Answer: Yes, Digital formats carry more data per second than analog formats

Explanation: The transmission of data from one area (point A) to another area (point B) is refered to as a signal. This signals can be transmitted from point A to point B through two different medium and formats, namely; analog formats and the digital format.

In the analog format, the data is converted into electrical pulse before it's been sent while in digital, data is been converted into binary format representing two amplitude before been sent. Note that devices or machines are at point b (destination) which will collect this pieces of information an convert them into readable and interpretable sounds or image.

Analog signals are a bit old and there is limitation on the size of data that can be transmitted in this formats.

The digital signal is more modern and it carry more data per second compared to the Analog.

So digital formats can hold more data than the analog.

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What volume of solution must be added to 4.0 mol of NaCl to make a 1.2 M solution?

Aleksandr [31]

Answer:

$\boxed {\boxed {\sf 3.3 \ liters}}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity=\frac{moles \ of \ solute}{liters \ of \ solution}}$

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.

molarity= 1.2 mol/L
moles of solute= 4.0 mol
liters of solution =x

Substitute the values into the formula.

$1.2 \ mol/L = \frac{4.0 \ mol}{x}$

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.

$\frac {1.2 \ mol/L}{1}=\frac{ 4.0 \ mol}{x}$

$4.0 \ mol *1=1.2 \ mol/L *x$

$4.0 \ mol = 1.2 \ mol/L *x$

x is being multiplied by 1.2 moles per liter. The inverse of multiplication is division, so divide both sides by 1.2 mol/L

$\frac{4.0 \ mol}{1.2 \ mol/L} = \frac{1.2 \ mol/L *x}{1.2 \ mol/L}$

$\frac{4.0 \ mol}{1.2 \ mol/L}=x$

The units of moles (mol) will cancel.

$\frac{4.0 }{1.2 } \ L =x$

$3.33333333 \ L=x$

The original measurements both have 2 significant figures, so our answer must have the same. For the number we found, this is the tenths place.

The 3 in the hundredth place tells us to leave the 3 in the tenths place.

$3.3 \ L\approx x$

Approximately <u>3.3 liters of solution</u> are needed to make a 1.2 M solution with 4.0 moles of sodium chloride.

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3 years ago

Using your knowledge of reagents that react with alkenes, what would be a reagent that you could use to check and see if the hyd

igomit [66]

Answer:

An halogen addition reaction, particularly bromine addition, could be used to check if the hydrogenation has completed.

Explanation:

The aim is to find a way to check if the hydrogenation process of an alkene has completed. So the logic should be use a reaction that needs the participation of the double bound of the alkene, and therefore shouldn't take place in the process has finished.

A simple organic reaction is the halogen addition reaction, which occur between the halogen molecule and the double bond of the alkene. Basically, the pi electrons of the double bond attacks a relatively electrophilic halogen atom following a mechanism that leads to the addition of two halogen atoms to the double bond. As a consequence the alkene transforms into an haloalkane.

Also, a commonly used halogen is bromine, as it has a reasonable reactivity and it has red colour, which allows to monitor the progression of the reaction.

Taking all this in account, we can say that using a bromine addition reaction to the alkene it's a good option to check the completion of the hydrogenation.

Note that the bromine will react only if the double bond is present. So, if the hydrogenation has completed, the reaction won't occur. Using bromine will be the best option, as it is red coloured. That means that is we add bromine to an incomplete hydrogenation, as it will react with the alkene, the colour should disappear.

In other words, to check if the hydrogenation reaction has completed, we should add bromine and see what happens to the colour. If it doesn't change, then the hydrogenation reaction has completed.

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3 years ago

What is the percent composition by mass of oxygen in Ca(NO3)2 (gram-formula= 164 g/mol)?

Neko [114]

To find this, we will use this formula:

Molar mass of element
------------------------------------ x 100
Molar mass of compound

So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.

Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol

When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!

95.994
-----------
164.086

When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.

So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.

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4 years ago

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djverab [1.8K]

1,05+ 0,69 + 1,82 = 3,56g of mixture

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3 years ago

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geniusboy [140]

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3 years ago