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blondinia [14]
3 years ago
12

An element occurs naturally as a gas. It is also characterized by a very

Chemistry
1 answer:
Delvig [45]3 years ago
8 0
I believe it would be c

Please brainiest
You might be interested in
How many molecules of N204 are in 85.0 g of N2O4?
alukav5142 [94]

Answer:

5.56 × 10^23 molecules

Explanation:

The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)

Using mole = mass/molar mass

Molar mass of N2O4 = 14(2) + 16(4)

= 28 + 64

= 92g/mol

mole = 85.0/92

= 0.9239

= 0.924mol

number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23

= 5.56 × 10^23 molecules

4 0
3 years ago
What is the speed of light?can a object move in speed of light?​
timurjin [86]

Answer:

The speed of light is the speed at which light travels. No, an object cannot move at the speed of light.  

Explanation:

The speed of light is 186,000 miles per second. An object with mass cannot move at the speed of light since it would take an infinite amount of energy to achieve that velocity, since only massless particles can travel at the speed of light. Also, you would have to factor in air friction, meaning even if an object were to reach such high speeds, it would instantly disintegrate.

4 0
2 years ago
How are a flashlight and lightening similar and how are they different?
grandymaker [24]
<span>A flashlight is an electric-powered light source; the light source is a light bulb or an LED. The electrical energy is converted into visible light. Flashlights can be hand-held or mounted to a platform. Light from a lighting, on the other hand, is formed by exciting electrons to a higher state. </span>
5 0
3 years ago
Please help with 3! Please give only the correct answer...
cupoosta [38]
The answer is:  " 1.75 * 10 ^(-10)  m " .
_________________________________________________________
Explanation: 
_________________________________________________________
This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
_________________________________________________________
Given: " 0.000000000175 m " ;  write this in "scientific notation.
_________________________________________________________
Note:   After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
                     There are NINE (9) zeros, followed by "175"
_______________________________________________________
To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding";  that is a decimal point with zeros PRECEDING an "integer expression"<span>
</span><span> (the "integer" is "175").</span>
______________________________________________________
We then take the "integer expression" (whatever it may be:  12, 5, 30000001 ; or could be a negative value,  etc.) ;  

→  In our case, the "integer expression" is:  "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit;  if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit);  and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case:  "175" ; becomes:  " 1.75" .
__________________________________________________
Then we write:  "  * 10^ "
__________________________________________________
   {that is "[times]"; or "multiplied by" :    [10 raised exponentially to the power of  <u>     </u> ]._____________________________________________________
 And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros";  if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write:  " * 10⁰ " ;    since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing:  " * 1 " .

If there are "trailing zeros" AND/OR or  any number of decimal places,  to the "right" of this expression; the combined number of spaces to the right is: 
  { the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case:  we have:  0.000000000175 * 10^(-10) .

Note:  The original notation was:

             →  " 0.000000000175 m "

{that is:  "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
____________________________________________________
So we have:
         →   " 0.000000000175 m " ;

Think of this value as:

        " 0. 0000000001{pseudo-decimal point}75   m ".

And count the number of decimal spaces "backward" from the
      "pseudo-decimal point" to the actual decimal; and you will see that there are "10" spaces (to the left).   
______________________________________________________
Also note:  We started with "9 (nine)" preceding "zeros" before the "1" ;  now we are considering the "1" as an "additional digit" ;
             →  "9 + 1 = 10" .
______________________________________________________
Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
 "NEGATIVE TEN" { "-10" } .

So we write this value as:  " 1.75 * 10^(-10)  m " .  

{NOTE:  Do not forget the units of measurement; which are "meters" —which can be abbreviateds as:  "m" .} . 
______________________________________________________
The answer is:  " 1.75 * 10^(-10)   m " .
______________________________________________________
4 0
3 years ago
Lead has a density of 11.3 g/mL. If you have 105 grams of lead,
Elza [17]

Answer:

<h2>Volume = 9.29 mL</h2>

Explanation:

Density of a substance can be found by using the formula

Density( \rho) =  \frac{mass}{volume}

From the question

Density = 11.3 g/mL

mass = 105 g

Substitute the values into the above formula and solve for the volume

That's

11.3 =  \frac{105}{v}

Cross multiply

11.3v = 105

Divide both sides by 11.3

v =  \frac{105}{11.3}

v = 9.29203

We have the final answer as

<h3>Volume = 9.29 mL</h3>

Hope this helps you

6 0
3 years ago
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