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olchik [2.2K]
2 years ago
14

In which groups of the modern periodic table are very active metals and very active non-metals placed?​

Chemistry
1 answer:
eimsori [14]2 years ago
5 0

Answer:

group 1 and are called Alkali metals. Similarly, very active non-metals are placed in group 17

Explanation:

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Is. Fe(s) ---------------&gt; Fe2+(aq) + 2e−<br>oxidation or reduction?​
frez [133]

Answer:

Oxidation occurs at the anode: Fe(s) | Fe2+(aq) half cell. (ii) Reduction occurs at the cathode: Ag(s) | Ag+(aq) half cell. Oxidation occurs at the anode: Pt | Sn2+(aq), Sn4+(aq) half cell. (iii) Electrons flow from the anode to the cathode: from the Pt(s) → Ag(s) electrode.

4 0
2 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
Pls help asap pls !!
klio [65]

Answer:

basic solution.

Explanation:

has a pH of 12.

it turns clear indicator pink.

it contains hydroxide ions which conduct electricity

3 0
2 years ago
Matter is never created or destroyed.
kumpel [21]
I believe it is true!
8 0
3 years ago
A car rolls down a ramp. What is the force acting on the car that causes the movement down the ramp?
Komok [63]
The answer to your question is gravity. you are welcome. :-D
6 0
3 years ago
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