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Nana76 [90]
3 years ago
10

A sample of iron has the dimension of 2cm x 3cm x2cm. If the mass of this rectangular shaped objects is 94g, what is the density

?
Chemistry
1 answer:
g100num [7]2 years ago
7 0
Volume iron = 2 x 3 x 2 => 12 cm³

Mass = 94 g

D = m / V

D = 94 / 12

D = 7.833 g/cm³

hope this helps!
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1. How many joules of heat are required to raise the temperature of 750 g of water from 11.0 oC to 19.0 oC?
Leya [2.2K]

Answer:

  1. 25080 J
  2. 146.9 g
  3. 92.58 °C
  4. 0.808 J/g°C
  5. 117.09 g
  6. a. 1708.8 kJ  b.1246.56 kJ
  7. 368.55 kJ
  8. 6.81 kJ
  9. 5.50 grams of methane produces more heat than 5.5 grams of propane.

Explanation:

  1. The specific heat capacity of water=4.18 J/gK

The enthalpy change is calculated using the formula: ΔH=MC∅ where ΔH is the change in enthalpy, M the mass of the substance, C the specific heat capacity of the substance and ∅ the temperature change.

Thus, ΔH= 750g × 4.18 J/gK × (19-11)K

=25080 J

2. Enthalpy change= mass of substance × specific heat capacity of the substance× Change in temperature.

ΔH= MC∅

M= ΔH/(C∅)

Substituting for the values in the question.

M=8750 J/(0.9025/g°C×66.0 °C)

=146.9 grams

3. Enthalpy change =mass × specific heat capacity × Temperature

ΔH= MC∅

∅ = ΔH/(MC)

=6500 J/(250 g × 4.18 J/g°C)

=6.22° C

Final temperature =98.8 °C - 6.22°C

=92.58 °C

4. Specific heat capacity =mass × specific heat capacity × Temperature change.

ΔH=MC∅

C= ΔH/(M∅)

Substituting with the values in the question.

C = 4786 J/(89.0 g×(89.5° C-23°C))

=0.808 J/g°C

5. Heat lost lost copper is equal to the heat gained by water.

ΔH(copper)= ΔH(water)

MC∅(copper)=MC∅(water)

M×0.385 J/g°C× (75.6°C- (19.1 °C+5.5°C))=100.0g×4.18 J/g°C×5.5 °C

M=(100.0g×4.18J/g°C×5.5°C)/(0.385 J/g°C×51 °C)

=117.09 grams.

6 (a). From the equation 1 mole of methane gives out 890.4 kJ

There fore 2 moles give:

(2×890.4)/1= 1780.8 kJ  

(b) 22.4 g of methane.

Number of moles= mass/ RFM

RFM=12 + 4×1

=16

No. of moles =22.4 g/16g/mol

=1.4 moles

Therefore 1.4 moles produce:

1.4 moles × 890.4 kJ/mol=

=1246.56 kJ

7. From the equation, 2 moles of aluminium react with ammonium nitrate to produce 2030 kJ

Number of moles = mass/RAM

Therefore 9.75 grams = (9.75/26.982) moles of aluminium.

=0.3613 moles.

If 2 moles produce 2030 kJ, then 0.3613 moles produce:

(0.3631 moles×2030 kJ)/2

=368.55 kJ

8. From the equation, 4 moles of ammonia react with excess oxygen to produce 905.4 kJ of energy.

Number of moles= mass/molar mass

RMM= 14+3×1= 17

Therefore 0.5113 grams of ammonia = (0.5113 g/17g/mole) moles

= 0.0301 moles

If 4 moles produce 905.4 kJ, then 0.0301 moles produce:

(0.0301 moles×905.4 kJ)/4 moles

=6.81 kJ

9. From the equations, one mole of methane produces 890 kJ of energy while one mole of propane produces 2043 kJ.

Lets change 5.5 grams into moles of either alkane.

Number of moles= Mass/RMM

For propane, number of moles= 5.5g/ 44.097g/mol

=0.125 moles

For methane number of moles =5.5 g/ 16g/mol

=0.344 moles

0.125 moles of propane produce:

0.125 moles×2043 kJ/mol

=255.375kJ

0.344 moles of methane produce:

0.344 moles× 890 kJ/mol

= 306.16kJ

Therefore, 5.5 grams of methane produces more heat than 5.5 grams of propane.

6 0
3 years ago
The basic structure of an amino acid contains the amine group, NH2, and a carboxylic acid group separated by one or more carbon
OLEGan [10]

The answer would be True

3 0
3 years ago
By measuring the distance from the crest of one wave to the crest of another we will know the _____ of the light.
stepan [7]
The wave length of light. This determinds the color/type of light.
4 0
2 years ago
Cavendish’s name for hydrogen gas was "inflammable air." The word inflammable means "to burn. "Why did Cavendish use this name?
saul85 [17]
He used that name because if you lit hydrogen on fire it would catch. therefore, it was dangerous to be around. think of it as a kind of warning. :) hope that helps.
8 0
3 years ago
Read 2 more answers
The equation for the combustion of CH4 (the main component of natural gas) is
Lera25 [3.4K]

Heat produced =  -13588.956 kJ

<h3>Further explanation</h3>

Given

The reaction of combustion of Methane

CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ

271 g of CH4

Required

Heat produced

Solution

mol of 271 g CH₄ (MW=16 g/mol0

mol = mass : MW

mol = 271 : 16

mol = 16.9375

So Heat produced :

= mol x ΔH°rxn

= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ

6 0
2 years ago
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