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denis23 [38]
2 years ago
5

If you add the same amount of heat to two different substances, will they both change phase?

Chemistry
1 answer:
lesantik [10]2 years ago
8 0

Answer:

It depends on their melting and/or their boiling points, because the heat provides the particles with kinetic energy to break the electrosatic bonds in the substances, which can differ in strength

Explanation:

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What is the empirical formula of a compound that is 18.8% Li, 16.3% C , and 64.9% O
zlopas [31]

Greetings!

To find the empirical formula you need the relative atomic mass of each element!

Li = 6.9

C = 12

O = 16

You can simply change the percentages into full grams

Li = 18.8g

C = 16.3g

O = 64.9


Then you use this to find the Number of moles = amount in grams / atomic mass

Li = 18.8 ÷ 6.9 = 2.7246

C = 16.3 ÷ 12 = 1.3583

O = 64.9 ÷ 16 = 4.0562

Then divide each number of moles by the smallest value:

Li = 2.7246 ÷ 1.3583 = 2.0

C = 1.3583 ÷ 1.3583 = 1

O = 4.0562 ÷ 1.3583 = 2.9 ≈ 3


So that means that there are 2 Li, 1 C, and 3 O

Empirical formula would be:

Li₂CO₃


Hope this helps!

3 0
3 years ago
10. When a 13.6 g sample of a compound containing only magnesium and oxygen is decomposed, 5.4 g of oxygen is obtained. What is
irinina [24]

Answer:5.4 g / 13.6 g *100

Explanation:Its is the correct answer

7 0
3 years ago
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Marianna [84]

Answer:

Under the concept of popular sovereignty, the people of each territory would decide whether or not slavery would be permitted.

Explanation:

hopes this help 3> D:

6 0
3 years ago
The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th
kirill115 [55]

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound  present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound  = 1570 seconds

Plugging in the values, we have  

Nt/No = (1/2)^(4710s/1570s)

Nt/No = (1/2)^3

Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

5 0
2 years ago
Compute the values of the diffusion coefficients for the interdiffusion of carbon in both α-iron (BCC) and γ-iron (FCC) at 900°C
bogdanovich [222]

Answer:

α-iron (BCC) has faster diffusion rate because of lower values in activation energy and pre-exponential value.

Explanation:

Taking each parameters or data at a time, we can determine the values/a constant for each parameters in the diffusion coefficient equation.

For α-iron (BCC), the diffusion coefficient = pre-exponential value,Ao × e^( -Activation energy,AE)/gas constant,R × Temperature.

Converting the given Temperature, that is 900°C to Kelvin which is equals to 1173.15K.

For α-iron (BCC), the pre-exponential value, Ao = 1.1 × 10^-6, and the activation energy, AE = 87400.

Thus, we have that the diffusion coefficient = 1.1 × 10^-6 × e(-87400)/1173.15 × 8.31.

Diffusion coefficient for α-iron (BCC) = 1.41 × 10^-10 m^2/s.

Also, For the γ-iron (FCC), the pre-exponential value, Ao = 2.3 × 10^-5 and the activation energy, AE = 148,00.

From these values we can see that both the exponential value, Ao and the activation energy for γ-iron (FCC) are higher than that of α-iron (BCC).

Thus, the diffusion coefficient for the γ-iron (FCC) = 2.3 × 10^-5 × e ^-(14800)/8.31 × 1173.15.

Then, the diffusion coefficient for the γ-iron (FCC) = 5.87 × 10^-12 m2/s.

Therefore, there will be faster diffusion in α-iron (BCC) because of lower activation energy and vice versa.

6 0
3 years ago
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