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Schach [20]
3 years ago
6

Imagine that the moon’s orbit wasn’t elliptical and it that it did not orbit the earth at a 5 degree angle. How would this chang

e the predictability of an eclipse happening every month?
Chemistry
2 answers:
zimovet [89]3 years ago
7 0

Lunar eclipse will happen each month, because of this.

Readme [11.4K]3 years ago
4 0

Answer:

Without the tilt, lunar eclipse will happen every month.

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Calculate the volume which 1.00 mole of a gas occupies at 1 atm and 298K?
elena-14-01-66 [18.8K]

Answer:

25.45 Liters

Explanation:

Using Ideal Gas Law PV = nRT => V = nRT/P

V = (1mole)(0.08206Latm/molK)(298K)/(1atm) = 25.45 Liters

7 0
3 years ago
Write chemical equation describing the complete combustion of liquid octane c8h18
Hunter-Best [27]
Ljiskdhf;hsdhsoh=2 lol
4 0
3 years ago
What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?
alina1380 [7]

Mass of H₂ needed to react with O₂ : 1.092 g

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Reaction

O₂(g) + 2H₂(g) → 2H₂O(g)

mass of O₂ : 8.75 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{8.75}{32}=0.273

From the equation, mol ratio of O₂ : H₂ = 1 : 2, so mol H₂ :

\tt \dfrac{2}{1}\times 0.273=0.546

Mass H₂ (MW=2 g/mol) :

\tt 0.546\times 2=1.092~g

4 0
3 years ago
Read 2 more answers
If 10.62 mL of a standard 0.3330 M KOH solution reacts with 98.20 mL of CH3COOH solution, what is the molarity of the acid solut
Nikolay [14]

Answer:

0.036 M of CH_{3} COOH

Explanation:

It is an example of acid-base neutralization reaction.

KOH  + CH_{3} COOH  ----> CH_{3} COO^{-} K^{+}   +   H_{2}O

Base           Acid                           Salt                                    

When two component react then the number of moles of both the component should be same, therefore the number of moles and acids and bases should be the same in the following .

Molarity= \frac{\textrm{No. of Moles}}{\textrm{Volume of the Particular Solution}}

No.of moles= Molarity × Volume of the Particular Solution

Therefore,

M_{1}V_{1} =M_{2}V_{2}------------------------------(1)

where

M_{1}= Molarity of Acid

V_{1}= Volume of Acid

M_{2}= Molarity of Base

V_{2}= Volume of Base

M_{1}=0.3330 M

V_{1}=10.62 mL

V_{2}=98.2 mL

M_{2}=??(in M)

Plugging in Equation 1,

0.3330 × 10.62 =M_{2}  × 98.2  

M_{2}=\frac{0.3330*10.62}{98.2}

M_{2}=0.036 M

3 0
3 years ago
A sample has a mass of 35.4g and a volume of 36.82mL. What is the density of the sample
vredina [299]

The density of the sample is 0.96 g/mL.


Density is found by dividing the mass of an object by its volume.


D=m÷v

D=35.4 g÷36.82 mL

D=0.96 g/mL

6 0
3 years ago
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