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spin [16.1K]
3 years ago
11

Question #2, thanks :)

Mathematics
1 answer:
tiny-mole [99]3 years ago
8 0

the first rate is 1.25 ur on your own for the rest

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A gas measuring 525 mL is collected at 104.66 kPa. What volume does this gas
Olegator [25]

Recall the ideal gas law:

<em>P V</em> = <em>n R T</em>

where

<em>P</em> = pressure

<em>V</em> = volume

<em>n</em> = number of gas molecules

<em>R</em> = ideal gas constant

<em>T</em> = temperature

If both <em>n</em> and <em>T</em> are fixed, then <em>n R T</em> is a constant quantity, so for two pressure-volume pairs (<em>P</em>₁, <em>V</em>₁) and (<em>P</em>₂, <em>V</em>₂), you have

<em>P</em>₁ <em>V</em>₁ = <em>P</em>₂ <em>V</em>₂

(since both are equal to <em>n R T </em>)

Solve for <em>V</em>₂ :

<em>V</em>₂ = <em>P</em>₁ <em>V</em>₁ / <em>P</em>₂ = (104.66 kPa) (525 mL) / (25 kPa) = 2197.86 mL

3 0
3 years ago
Plot the data for the functions f(x) and g(x) on a grid
wariber [46]
The graph (by some miracle) has been uploaded for you. It is just about the first time I've done this sort of thing, and I've answered nearly 800 questions. 

The first thing you have to do is study the graph. The two functions are
f(x) = 4^x That's the curved graph. (in red)
g(x) = x + 4. That's the straight line. (in blue)

You know that the first one is not a linear relationship because the x values go from integer values -2 to 2 (including 0). The y values are a bit different. They go from 1/16 to 16 with those integer values. So you could try y = 4^(-x). It doesn't work, but you could try it. It gives the table numbers for y in the reverse order that the table you are given goes. For x you get -2 -1 0 1 2 and for y you would get 16 4 1 1/4 1/16.

You could try y = (1/4)^x
For this try, you would get x = -2 -1 0 1 2 and for y = 16 4 0 1/4 and 1/16 
but that doesn't work either.

You could try until you get y = 4^x which does work.

g(x) is a lot easier to deal with. It looks better behaved. as x goes up, so does y. You will find that the y values obey y = x + 4. You could try other lines, but that one works. Many times it's just a guess

7 0
3 years ago
What is the value of x? <br><br> A. 3cm <br> B. 4cm<br> C. 5cm<br> D. 4.5cm
Alexxx [7]

Answer:

5 cm

Step-by-step explanation:

32 / 4 = 8

40 / 8 = x

40 / 8 = 5

32 and 4 are corresponding, and 40 and x are as well. Find the relation between 32 and 4, then apply it to 40 to find the value of x.

8 0
3 years ago
I FORGOT HOW TO DO THIS I NEED HELP ASAP!!!
Sidana [21]
Slope form: y=mx + b
slope is 2/5, fill into equation
Y = 2/5x + b
Plug in point to find b
1 = 2/5(0) + b
1 = 0 + b, b = 1
The answer is A. Y = 2/5x + 1
8 0
3 years ago
Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7
photoshop1234 [79]
Hope I can be of assistance!

3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \  \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}
3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \  3b^{-5}b^2c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  b^{-5}b^2=\:b^{2-5}=\:b^{-3}
3b^{-3}c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  c^{-4}c^7=\:c^{-4+7}=\:c^3
3b^{-3}c^3

\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \  b^{-3}=\frac{1}{b^3} \ \textgreater \  3\cdot \frac{1}{b^3}c^3

\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{1\cdot \:3c^3}{b^3}

Finally
\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \  \frac{3c^3}{b^3}

Hope this helps!
8 0
4 years ago
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