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3 years ago

Calculate the mass (in g) of 2.1 x 1024 atoms of W.

1 answer:

6 0

Set up your units to algebraically cancel out the original unit, and end up with the units of the answer in the final numerator. We'll first convert to moles by using Avogadro's number, then the atomic mass of tungsten (W) from the periodic table, since the mass, in grams, of one mole of any element is the atomic mass number:
2.1x10∧24 atoms W x (1 mole/6.022x10∧23 atoms) x (183.84g/1 mole) = 641.089 grams
Since your given number has only two significant figures (2.1), your answer should also only express in 2 sig figs: 6.4x10∧2 grams

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In the first 85.0 s of this reaction, the concentration of no dropped from 1.12 m to 0.520 m . calculate the average rate of the

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In the first 85.0 s of this reaction, the concentration of no dropped from 1.12 m to 0.520 m .

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The speed at which a chemical reaction takes place is the rate of the reaction. It is the concentration change per unit time of a reactant in a reaction.

Since the concentration of NO reduces to half its initial concentration in 85 seconds that is from 1.12m to 0.520m, it can be said that 85 seconds is the half life interval for the reaction, Hence on average, half reaction is completed in the time interval of 85 seconds.

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2 years ago

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4 years ago

The flow of energy through living systems can be modeled as food chains, food webs, or energy pyramids. The energy organisms nee

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3 years ago

The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves

adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

Where ΔT is change in temperature = 0.400°C

Kf is freezing point constant of the solvent = 1.86°C/m

m is molality of the solution (Moles of solute / kg of solvent)

And i is Van't Hoff constant (1 for a nonelectrolyte)

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

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3 years ago