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kolezko [41]
2 years ago
6

Ordinary water boils at 100°C. Can it be made to boil at 95°C or 105°C.​

Chemistry
2 answers:
zhuklara [117]2 years ago
7 0

The best way is add alcohol .

The common way is that decrease the atmospheric pressure .

This will work as per Boyles law ,and boiling point can be decreased.

but it has few restrictions

  • You can't decrease pressure simply at normal conditions
  • For decreasing pressure you have to go hill tops .

Hence there is a spare way

  • Add alcohol

General alcohols like ethanol,butanol can decrease the boiling point of water

slava [35]2 years ago
4 0

It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.

<h3>Concept :</h3>

To boil water at 95°C, decrease the atmospheric pressure.

At 105°C, the water will be boiling quickly than normal at 100°C.

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Convert 88.2°C into °F
prisoha [69]
The formula for converting Celsius to Fahrenheit is:
Let Celsius be C

(C * 1.8) + 32 = the value in Fahrenheit.

(88.2*1.8) + 32 = 190.76 degrees Fahrenheit.

Hope this helps!
5 0
3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
In a particular redox reaction, no−2no2− is oxidized to no−3no3− and fe3 fe3 is reduced to fe2 fe2. complete and balance the equ
anyanavicka [17]

Balanced chemical equation is  -  3Fe2+ + NO3- + 4H+ → 3Fe3+ + NO + 2H2O

Fe2+ + NO3- + H+ → Fe3+ + NO3- + NO + H2O

Fe+22+ + N+5O-23- + H+1+ → Fe+33+ + N+5O-23- + N+2O-2 + H+12O

3Fe2+ + NO3- + 4H+ → 3Fe3+ + NO + 2H2O

A balanced chemical equation specifies the quantities of reactants and products required to meet the Law of Conservation of Mass. This signifies that there is the same number of each sort of atom on the left side of the equation as there are on the right side. The rule of a balanced chemical equation is used to determine if two weights put on opposing sides of the fulcrum will balance each other.

To learn more Balanced chemical equations about please visit -
brainly.com/question/15052184
#SPJ4

3 0
2 years ago
Find the mass of an object on earth if its weight is 100N
Kaylis [27]

Answer:

0.19 kg

8 N

1.964 N

58.86 N

15 kg

Explanation:

w = Weight of an object = 100 N

g = Acceleration due to gravity of Earth = 9.81\ \text{m/s}^2

m = Mass of an object

Weight is given by

w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{100}{9.81}\\\Rightarrow m=10.19\ \text{kg}

Mass of the object is 0.19 kg

g = 2\ \text{m/s}^2

m = 4 kg

w=mg\\\Rightarrow w=4\times 2\\\Rightarrow w=8\ \text{N}

Weight of the object is 8 N.

m = 200 g

w=mg\\\Rightarrow w=0.2\times 9.81\\\Rightarrow w=1.964\ \text{N}

Weight of the object is 1.964 N.

Weight is the force the Earth exerts on an object which is on the surface of the Earth.

m = 6kg

w=mg\\\Rightarrow w=6\times 9.81\\\Rightarrow w=58.86\ \text{N}

The force of the object is 58.86 N.

w = 150 N

a = g = 10\ \text{m/s}^2

m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{150}{10}\\\Rightarrow m=15\ \text{kg}

Mass of the object is 15 kg.

7 0
3 years ago
Which equation is an example of a redox reaction?
VikaD [51]
I think is D,,,,,,,,,,
8 0
2 years ago
Read 2 more answers
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