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nlexa [21]
3 years ago
5

Who pass the grade quarter?.

Chemistry
2 answers:
Fed [463]3 years ago
8 0

Answer:

me

Explanation:

ella [17]3 years ago
8 0

Answer:

If you're doing good for an assignment and quiz, then you can pass the grade quarter.

Explanation:

You might be interested in
WHAT DID YOU OBSERVE HAPPENED WHEN AN S-WAVE TRAVELED INTO A LIQUID?
alexgriva [62]
It will spread out making the sound travel faster then it usually does in the air(i think) let me know if you need help with anything else
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
g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0
2 years ago
What causes attraction and repulsion between atoms?​
Jobisdone [24]

Answer:

Repulsive forces exist only when atoms are very close to each other. (3/14) "They [the atoms] will approach until both nuclei will simply shove each other because both of them are positive." The balance between the attraction and repulsion forces determines how close the atoms can get. The relationships between the magnitude and direction of repulsive and attractive forces. A stable state of a bond is when attractive forces balance repulsion forces. “A stable state between two atoms is when they attract each other with a force that equals the force that they repel each other.”

7 0
3 years ago
Which would have a stronger attraction to the nucleus, the outer electrons of a carbon atom, or the outer electrons of an oxygen
Arte-miy333 [17]

Answer: Outer electrons because they revolve around the nucleus and can be multiple more than one

Explanation:

Hope it helps you you

7 0
3 years ago
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