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eduard
3 years ago
10

Write this numeral in scientific notation. 820.5

Chemistry
1 answer:
Viefleur [7K]3 years ago
6 0
The numeral in scientific notation would be 8.205x10^-2
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How many grams of carbon are in 0.24 moles of carbon?
Ronch [10]

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I'm converting this if I could remember how

2.882568

2 110321/ 125000

T-T sorry if I'm wrong I have bad memory

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3 years ago
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Determine all values of hh and k for which the system has no solution
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6 0
3 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
3 years ago
Describe the process of making the four major macromolecules.
dybincka [34]

Answer:

<em>Proteins, carbohydrates, nucleic acids, and lipids are the four major classes of biological macromolecules—large molecules necessary for life that are built from smaller organic molecules. Macromolecules are made up of single units known as monomers that are joined by covalent bonds to form larger polymers.</em>

Explanation:

Have a great day!

8 0
3 years ago
Read 2 more answers
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