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3 years ago

We can’t see the dark side of the moon from earth because

1 answer:

3 0

On the Moon, there is (as on Earth) day and night (though the day is as long as month onEarth) and no spot remains dark for longer (except for the pole caps as on Earth). There is, of course, the far side of the Moon, most of which is invisible from Earth.

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You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?

umka21 [38]

Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

$N=\frac{eq_{solute}}{V_{solution}}$

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

$eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq$

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

$m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4$

Best regards.

4 0

3 years ago

A solution is made by mixing equal masses of methanol, CH 4 O , and ethanol, C 2 H 6 O . Determine the mole fraction of each com

Fynjy0 [20]

Answer:

Mole fraction of $CH_4O$ = 0.58

Mole fraction of $C_2H_6O$ = 0.42

Explanation:

Let the mass of $CH_4O$ and $C_2H_6O$ = x g

Molar mass of $CH_4O$ = 33.035 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}$

$Moles_{CH_4O}=\frac{x}{33.035}\ mol$

Molar mass of $C_2H_6O$ = 46.07 g/mol

Thus,

$Moles= \frac{x\ g}{46.07\ g/mol}$

$Moles_{C_2H_6O}=\frac{x}{46.07}\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}$

$Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58$

Mole fraction of $C_2H_6O$ = 1 - 0.58 = 0.42

6 0

3 years ago

How many molecules are in 3 moles of potassium bromide (KBr)

sattari [20]

Answer:

Your strategy here will be to use the molar mass of potassium bromide,

KBr

, as a conversion factor to help you find the mass of three moles of this compound.

So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations,

, and bromide anions,

−

. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.

Use the periodic table to find the molar masses of these two elements. You will find

For K:

39.0963 g mol

−

For Br:

79.904 g mol

−

To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements

M KBr

39.0963 g mol

−

79.904 g mol

−

≈

119 g mol

−

So, if one mole of potassium bromide has a mas of

119 g

m it follows that three moles will have a mass of

moles KBr

⋅

molar mass of KBr



119 g

mole KBr

357 g

You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

mass of 3 moles of KBr

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

360 g

∣

−−−−−−−−−

Explanation:

answer: 360 g 

6 0

3 years ago

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

Alexxandr [17]

Specific heat capacity is the required amount of heat per unit of mass in order to raise teh temperature by one degree Celsius. It can be calculated from this equation: H = mCΔT where the H is heat required, m is mass of the substance, ΔT is the change in temperature, and C is the specific heat capacity.

H = mCΔT
2501.0 = 0.158 (C) (61.0 - 32.0)

C = 545.8 J/kg·°C

5 0

3 years ago

The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is -5,074 kJ/mol. This value is the Delta. Hrxn f

Inessa [10]

Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is $\Delta H_{\rm rxn}$ for $\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O$ .

<h3>What is the enthalpy of reaction?</h3>

The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.

The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.

The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.

The combustion of octane can be given as:

$\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O$

Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is $\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O$ . Thus, option B is correct.

Learn more about enthalpy of reaction, here:

brainly.com/question/1657608

6 0

2 years ago