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3 years ago

7

What is one advantage of using a computer over a graphing calculator

2 answers:

Arlecino [84]3 years ago

5 0

Answer:

one advantage of using a computer over a graphing calculator is

Computer hold larger amounts of data is the correct answer

Explanation:

There are several advantages of using a computer over a graphing calculator the important one is that the computer holds a larger amount of data and calculators do not because it has very short space for memory storage.

by using the computer, the complex calculation can be done rapidly.

CaHeK987 [17]3 years ago

3 0

PCs are usually more powerful devices w/ higher resolutions than the common graphing calculator.

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The relationship between molecular velocities and temperature is a _____ relationship.

elixir [45]

The correct answer that would best complete the given statement above would be option 2. <span>The relationship between molecular velocities and temperature is a direct relationship. In other words, their relationship is directly proportional. Hope that this is the answer that you are looking for. </span>

3 0

3 years ago

Read 2 more answers

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

A metal sample has a mass of 64.2 g. What is the density of this metal?

brilliants [131]

density= mass/volume

we need the volume of the metal to find the density, in which case the answer would be 64.2/volume=density

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3 years ago

3. Why is timber considered a renewable

sergejj [24]

Answer:

B

Explanation:

4 0

3 years ago

What is the mass of 1.2 x 1023 atoms of arsenic?

Gre4nikov [31]

Answer:

14.93 g

Explanation:

First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:

1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As

Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:

0.199 mol * 74.92 g/mol = 14.93 g

Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.

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3 years ago