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Scorpion4ik [409]
2 years ago
15

A 23.7 g piece of iron at 54.9°C is cooled and releases 338 J of heat. The specific heat of iron is 0.450 J/g°C.

Chemistry
2 answers:
ohaa [14]2 years ago
8 0

Answer:

The heat capacity and the specific heat are related by C=cm or c=C/m. The mass m, specific heat c, change in temperature ΔT, and heat added (or subtracted) Q are related by the equation: Q=mcΔT. Values of specific heat are dependent on the properties and phase of a given substance.

Explanation:

olya-2409 [2.1K]2 years ago
5 0

Answer:

23.2 degrees c

Explanation:

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In another experiment, if 80 xo3 molecules react with 104 brz3 molecules how many br2 molecules will be produced which reactant
BaLLatris [955]

This is an incomplete question, here is a complete question.

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

In another experiment, if 80 XO_3 molecules react with 104 BrZ_3 molecules. How many Br_2 molecules will be produced which reactant will be used up in the reaction.

Answer : The number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

Explanation :

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

First we have to determine the limiting reagent.

From the balanced reaction we conclude that,

As, 8 molecules of BrZ_3 react with 6 molecule of XO_3

So, 104 molecules of BrZ_3 react with \frac{104}{8}\times 6=78 molecule of XO_3

From this we conclude that, XO_3 is an excess reagent because the given moles are greater than the required moles and BrZ_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the molecules of Br_2

From the reaction, we conclude that

As, 8 molecules of BrZ_3 react to give 4 molecules of Br_2

So, 104 molecules of BrZ_3 react to give \frac{104}{8}\times 4=52 molecules of Br_2

Hence, the number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

4 0
3 years ago
How much heat is absorbed when 63.7 g H2O(l) at 100 degrees Celsius and 101.3kPa is converted to steam at 100
Alona [7]

Answer:

Q = 143,921 J = 143.9 kJ.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

Q=m*\Delta _{vap}H

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

Q=63.7g*2259.36J/g\\\\Q=143,921J=143.9kJ

Regards!

3 0
3 years ago
Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed vrms for diatomic oxygen at 50∘C
trapecia [35]

Answer:(16)(2000)m/s=32000m/s

Explanation: A diatomic oxygen has a molar mass of 16

8 0
3 years ago
What does the chemical formula CaCl, show about the compound it represents? It is made up of one element. It is made up of two e
zavuch27 [327]

Answer:

It is made up of two elements.

Explanation:

To answer the question given above,

We shall determine the number of elements present in CaCl₂. This can be obtained as follow:

CaCl₂ contains calcium (Ca) and chlorine gas (Cl₂).

This implies that CaCl₂ contains two different elements.

Now, considering the options given in the question above, CaCl₂ is made up of two elements.

5 0
2 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
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