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ryzh [129]
2 years ago
12

When does the composition of a substance does not change?

Chemistry
1 answer:
blagie [28]2 years ago
4 0
I think when it does not react with anything I think
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3. How many moles of silver is 8.46x1024 atoms of<br> silver?
ioda

Answer:

<h2>14.05 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{8.46 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 14.053156

We have the final answer as

<h3>14.05 moles</h3>

Hope this helps you

6 0
2 years ago
34.969amu)(0.7577) =<br> (36.966amu)(0.2423)
Natasha_Volkova [10]

Answer:

26.4960 is the answer for the first one

8.9569 is the answer for the second one

3 0
3 years ago
A geochemist examines a piece of metal that he found in the soil. He performs tests to identify the metal from its density, elec
aniked [119]

Answer:

what r the statements?

Explanation:

attach or write them

3 0
3 years ago
For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0
2 years ago
To convert from °F to °C: T(°C) = T(°F - 32) × 5/9
DaniilM [7]

Answer:

Your notation is a bit confusing, let me write it more clearly.

Explanation:

( Temperature in °F − 32) × 5/9 =  Temperature in °C

4 0
2 years ago
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