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Leno4ka [110]
3 years ago
5

Water will move from a ____________ salt solution to a ____________ salt solution when they are across a differentially permeabl

e membrane.
Chemistry
1 answer:
const2013 [10]3 years ago
6 0
The answers are low concentrated (dilute) and high concentrated respectively. 

As the low concentrated salt solution has a higher water potential than that of the high concentrated salt solution, water molecules will flow from the region of higher water potential to the region of lower water potential, thus from the dilute salt solution to the high concentrated salt solution. This is due to the movement called osmosis. Note that osmosis also requires water to flow through a differentially permeable membrane, which means the membrane can allow certain substances (not all) to go in or out. If the differentially permeable membrane is not present, the movement of water molecules may be regarded as diffusion. 

Therefore, the answers for the blanks are low concentrated and high concentrated.




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How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?
hoa [83]

Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

______

NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

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Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.
JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

  • H: 1.008;
  • Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

  • Hydrogen gas \rm H_2, and
  • Magnesium chloride, which is a salt.

The chemical equation will be something like

\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}],

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of \rm H_2 that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be \rm MgCl_2.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq).

Balance the equation. \rm MgCl_2 contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq).

The number of \rm Mg and \rm Cl atoms shall be the same on both sides. Therefore

\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq).

The number of \rm H atoms shall also conserve. Hence the equation:

\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq).

How many moles of HCl are available?

M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}.

\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol.

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of \rm HCl is 2 while the coefficient in front of \rm H_2 is 1. In other words, it will take two moles of \rm HCl to produce one mole of \rm H_2. \rm 4.52576\;mol of \rm HCl will produce only one half as much \rm H_2.

Alternatively, consider the ratio between the coefficient in front of \rm H_2 and \rm HCl is:

\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}.

\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol.

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as \rm 1.00\times 10^{5}\;Pa, that volume will be 22.7 liters.

V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L, or

V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L.

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

5 0
3 years ago
At standard ambient temperature and pressure (SATP), a gas has density of 1.5328g/L. What is the molar mass of the gas?
ahrayia [7]

The standard ambient temperature and pressure are

Temperature =298 K

Pressure = 1atm

The density of gas is 1.5328 g/L

density = mass of gas per unit volume

the ideal gas equation is

PV = nRT

P = pressure = 1 atm

V = volume

n = moles

R= gas constant = 0.0821 Latm/mol K

T = 298 K

moles = mass / molar mass

so we can write

n/V = density / molar mass

Putting values

Pressure=\frac{nRT}{V}=\frac{massXRT}{VXmolarmass}

Pressure=\frac{densityXRT}{molarmass}

molarmass=\frac{densityXRT}{Pressure}=\frac{1.5328X0.0821X298}{1}=37.50

Thus molar mass of gas is 37.50g/mol

6 0
4 years ago
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