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RUDIKE [14]
3 years ago
10

The atomic number is the number of protons in the atom of the element

Chemistry
1 answer:
ipn [44]3 years ago
5 0

Answer:

True

Explanation:

If you look closley at the nucleus, you don't count the neutrons just the prtons which then effect the electrons.

Good luck :)

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Which substance can be decomposed by a chemical change?<br> (1) Co (3) Cr(2) CO (4) Cu
kotykmax [81]
Answer: (2) CO

Co, Cr, and Cu are elements. Elements cannot be broken down by a chemical change, but compounds can. Since CO is a compound, it can be decomposed by a chemical change.
7 0
3 years ago
what additional detail supports the point on view in the chart? 1-it doesn’t make sense to have an “honor system.” 2-someone sho
Katena32 [7]

Answer:

I wanna say 3

Explanation:

Provincetown is known for them so getting rid of them would be a bad thing.

8 0
3 years ago
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
Whats the molar mass of CO
g100num [7]
28.01 g/mol

hope that helped
7 0
3 years ago
Read 2 more answers
A resting adult requires about 240 ml of pure oxygen/mind and breathes about 12 times every minute. if inhaled air contains 20 p
Dmitry_Shevchenko [17]

12 times breathe give 240 ml of pure O_{2}. Each breathe gives 20 ml of O_{2}.

Let us consider, volume of air per breathe= x ml.

Pure O_{2} from inhaled air= \frac{20}{100}x ml and Pure O_{2} from exhaled air= \frac{16}{100}x ml.

Pure O_{2} from inhaled and exhaled air= 20 ml

So, \frac{20}{100}x + \frac{16}{100}x = 20

Therefore, x = 55.5 ml

So, volume of air per breath= 55.5 ml.


7 0
3 years ago
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