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bogdanovich [222]
2 years ago
14

1. What is the specific heat (C) of an unknown sample that weighs 5.0 grams, absorbds 250.0j of heat and has a temperature

Chemistry
1 answer:
maks197457 [2]2 years ago
7 0
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C

i hope that this eg gonna help u
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Explanation:

to lose or gain e- an atom has to donate or receive e- to to this it requires the more easily it can lose an electron the more electro positive it is and the more easily it gains electron the more electro negative it is.

4 0
3 years ago
If there is currently 50kg of U-235 present in Oklo, how much must have been present 750 million years ago when the reaction too
Viefleur [7K]
To answer this question, you need to know the concept of half-life, which is how a radioactive material decreases in mass over time.

The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:

\frac{750}{703.8}  = 1.066
Now, take this scale factor and multiply it by the current mass, like this:

50 \times 1.066 = 53.3
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:

50 + 53.3 = 103.3
I hope this helped you. Fell free to ask any further questions.
7 0
3 years ago
9 How many grams of O₂ are needed to produce 15.5 g Fe₂O3 in the following reaction? Fe(s) + O₂(g) → Fe₂O3 (s)​
Slav-nsk [51]

Answer:

Explanation:

so u can work out the amount of moles in FeO3 by doing mr of fe3o3 is 55.8*3+16*3=215.4

moles= mass/mr so you do 15.5g/215.4=0.0719 moles

then using 1 to 1 ratio so O2 moles is 0.0719

then use the equation mass=mole*mr

so 0.0719*16=1.15g

hope this make sense :)

8 0
3 years ago
Question 1<br> 1 pts<br> How many mols of bromine are present in 35.7g of<br> Tin(IV) bromate?
sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol

So, there are 0.0814 moles of bromine in 35.7g of  Tin(IV) bromate.

3 0
3 years ago
HELP PLEASE I BEG OF YOU
Vitek1552 [10]

Answer:

37.7%

Explanation:

58% of 65 is 37.7%

7 0
2 years ago
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