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Murljashka [212]
3 years ago
8

What are tidal conditions in science terms

Chemistry
1 answer:
Anon25 [30]3 years ago
8 0
They are the tides and conditions of the oceans
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What are isoelectronic and isostructural species?
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Isoelectronic species are ions or elements that have equal number of electrons. From the root word, electron of the word isoelectronic. On the other hand, isotructural species are compounds with the same structures.Hope this answers the question. 
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Here is a more complex redox reaction involving the permanganate ion in acidic solution: 5fe2+ + 8h+ + mno4− → 5fe3+ +mn2+ + 4h2
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The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,
Katyanochek1 [597]

Answer:

a

No

b

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is P = 100 \  mmHg =  \frac{100}{760}=  0.13156 \ atm

The temperature of CHCl3 is T  =  283 \  K

The volume of the container is V_c =  380mL =  380 *10^{-3}\  L

The temperature of the container is T_c  =  283 \  K

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

n  =  \frac{m }{M }

Here M is the molar mass of CHCl3 with the value M  =  119.38 \ g/mol

=> n  =  \frac{ 0.380 }{119.38 }

=> n  =  0.00318 \  mols

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

n_g  =  \frac{PV}{RT}

Here R is the gas constant with value R =  0.08206 L \  atm /mol\cdot  K

So

          n_g  =  \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}

          n_g  =  0.00215 \  mols

Given that the number of moles of  CHCl3 evaporated is less than the number of moles of CHCl3  initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of  air so the pressure at equilibrium is  100 mmHg

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3 years ago
Given 40 grams of magnesium. How many mole is this
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To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.

We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'

\frac{40 g Mg}{1} x\frac{1 mole Mg}{24.305 g Mg}

In this, we can see that the unit 'grams' will cancel out to leave us with moles.

In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.

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