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aliya0001 [1]
3 years ago
5

In what way is refraction and reflection similar?

Chemistry
1 answer:
MatroZZZ [7]3 years ago
6 0
They are similar in that they both change lights path way.
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In which state of matter do particles have the least amount of energy
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solid

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Based on the graph below, if you chose an alien at random on day 20, what form would it most likely be in?
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4 0
3 years ago
Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooct
Aleksandr [31]

Answer:

1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈  +  25/2O₂  →   8CO₂  +  9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density . Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8)  = 2.55×10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles . molar mass

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂

If we convert to kg  1.12×10¹⁴ g / 1000 =  1.12×10¹¹ kg

6 0
3 years ago
6
Vanyuwa [196]

Answer:

If the volume of a gas increased from 2 to 6 L while the temperature was held constant, <u><em>the  pressure of the gas decreased by a factor of 3.</em></u>

Explanation:

Boyle's law that says "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or

P * V = k

To obtain the proportionality factor k you must make the quotient:

k=\frac{V2}{V1} =\frac{6 L}{2 L}

k= 3

This means that <u><em>if the volume of a gas increased from 2 to 6 L while the temperature was held constant, the  pressure of the gas decreased by a factor of 3.</em></u>

4 0
3 years ago
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