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3 years ago

When might a galaxy appear redshifted and why?

1 answer:

6 0

Redshift is said to be experienced when the light or any other electromagnetic radiations from an object undergo an increase in the wavelength but it does not mean that the light literally becomes red.

Redshift occurs due to the expansion of space. It means the farther away the galaxy is from us, the more space it has expanded in the time.

Because as the light left the galaxy, more of the light has been stretched and so more redshifted the light is.

You might be interested in

What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl?

Ainat [17]

Given 3.72 g of P and 21.28g of Cl, converting these to moles: ( 3.72 g P)(mol P/30.97 g P) = 0.12 mol P (21.28 g Cl)(mol Cl/35.45 g Cl) = 0.60 mol Cl P:Cl = 0.12/0.60, therefore P: Cl =1/5 Therefore, PCl5 hope it helps

4 0

3 years ago

A major component of gasoline is octane (C8H18). When liquid octane is burned in air it reacts with oxygen gas to produce carbon

Paraphin [41]

Answer:

The answer to your question is 0.4 moles of Oxygen

Explanation:

Data

Octane (C₈H₈)

Oxygen (O₂)

Carbon dioxide (CO₂)

Water (H₂O)

moles of water = ?

moles of Oxygen = 1

Balanced chemical reaction

C₈H₈ +10O₂ ⇒ 8CO₂ + 4H₂O

Reactant Element Products

8 C 8

8 H 8

20 O 20

Use proportions to solve this problem

10 moles of Oxygen ----------------- 4 moles of water

1 mol of Oxygen ------------------ x

x = (4 x 1) / 10

x = 4 / 10

x = 0.4 moles of water

7 0

3 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

A gas sample occupies 4.2L at a pressure of 101 kPa. What volume will it occupy if the pressure is increased to 235 kPa

taurus [48]

1.8 L

<h3>Explanation</h3>

The volume of a gas, V, is inversely proportional to the pressure on it, P. That is:

V₁ · T₁ = V₂ · T₂.

Rearranging gives:

V₂ = V₁ · T₁ / T₂ = 4.2 × (101 / 235) = 1.8 L

4 0

2 years ago

Help on 2 and 3 due tomorrow pls help

Luda [366]

Answer:

b= they are grouped differently, but all the atoms are still there.

4 0

2 years ago