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jasenka [17]
3 years ago
15

Iron has density 7.87 g/cm^3. if 52.4 g of iron is added to 75.0 mL of water in a graduated cylinder, to what volume reading wil

l the water level in the cylinder rise?
Chemistry
1 answer:
vlada-n [284]3 years ago
8 0

Answer:

6.66 mL

Explanation:

The increase in the volume is due to the addition of the iron whose volume can be calculated as:

Using,

Density = Mass / Volume

Given that:

Density of Iron = 7.87 g/cm³

Mass of iron = 52.4 g

Thus, volume is:

Volume = Mass / Density = 52.4 / 7.87 cm³ = 6.66 cm³

Also, 1 cm³ = 1 mL

<u>The rise in the volume = 6.66 mL</u>

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I need help with my homework.Calculate the number of formula units in: 41.5 grams CaBr2
lana66690 [7]

Answer:

\text{ }1.25\times10^{23}\text{ formula units}

Explanation:

Here, we want to calculate the number of formula units in the given molecule

We start by getting the number of moles

To get the number of moles, we have to divide the mass given by the molar mass

The molar mass is the mass per mole

The molar mass of calcium bromide is 200 g/mol

Thus, we have the number of moles as follows:

\frac{41.5}{200}\text{ = 0.2075 mol}

The number of formula units in a mole is:

1\text{ mole = 6.02 }\times10^{22}\text{ formula units}

The number of formula units in 0.2075 mole will be:

0.2075\text{ }\times6.02\times10^{23}\text{ = }1.25\times10^{23}\text{ formula units}

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Answer:

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3 years ago
The standard free-energy changes for the reactions below are given. Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/mol ATP → A
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Answer:

-12.5 kJ/mol

Explanation:

The free-energy predicts if a reaction is spontaneous or not. If it is, ΔG < 0. When a reaction happens by steps, the free-energy of the global reaction can be calculated by the sum of the free-energy of the steps (Hess law). If it's needed to operations at the reaction the same operation must be done in the value of ΔG (if the reaction is inverted, the signal of ΔG must be inverted).

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

ATP → ADP + Pi                             ∆G'° = –30.5 kJ/mol (x-1)

--------------------------------------------------------------------------------------

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

Pi + ADP → ATP                             ∆G'° = 30.5 kJ/mol

The bold compounds are in opposite sides, so they'll be canceled in the sum of the reactions:

Phosphocreatine + ADP → creatine + ATP

∆G'° = -43.0 + 30.5

∆G'° = -12.5 kJ/mol

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