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saw5 [17]
3 years ago
12

Why should we put a glass on candles for experiments and why should we put candles on sand trays ?​

Chemistry
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

1, you should use candles for experiments because, You created an area of low pressure! When the experiment is run you can see tiny bubbles escaping under the glass which shows that the increased air pressure from the heated air as the candle burns. Once the candle runs out of oxygen, the candle burns out and the remaining air inside cools down.

2. you should put candles on sand trays, Pour layers of sand into a glass vase, alternating colors with each layer. For a more interesting look, try uneven layers in various heights. Keep layering sand until you're a few inches from the top of the vase. Place your votive candle in the center of the sand.

Explanation:

its scientific proven

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Which statement explains why a C–O bond is
Nataly_w [17]

Answer: (3) The difference in electronegativity between  carbon and oxygen is greater than that  between fluorine and oxygen.

Explanation: Polarity of a molecule is due to the difference in electronegativity of the atoms. More is the electronegativity difference, more is the polarity.

Electronegativity of carbon = 2.5

Electronegativity of oxygen = 3.5

Electronegativity of fluorine = 4.0

Thus the difference in electronegativity of carbon and oxygen is=(3.5-2.5)= 1.0

Thus the difference in electronegativity of fluorine and oxygen is=(4.0-3.5)= 0.5.

Thus C-O bond is more polar than F-O bond.

4 0
3 years ago
Read 2 more answers
The dissolution of 0.200 l of sulfur dioxide at 19 °c and 745 mmhg in water yields 500.0 ml of aqueous sulfurous acid. The solut
aivan3 [116]

Answer:

Molarity=1.22\ M

Explanation:

Given:  

Pressure = 745 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 745 / 760 = 0.9803 atm

Temperature = 19 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (19 + 273.15) K = 292.15 K  

Volume = 0.200 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K  

⇒n = 0.008174 moles

From the reaction shown below:-

H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O

1 mole of H_2SO_4 react with 2 moles of NaOH

0.008174 mole of H_2SO_4 react with 2*0.008174 moles of NaOH

Moles of NaOH = 0.016348 moles

Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)

So,

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.016348}{0.0134}\ M

Molarity=1.22\ M

8 0
3 years ago
If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show
Talja [164]

Answer:

4.5 g/L.

Explanation:

  • To solve this problem, we must mention Henry's law.
  • Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
  • It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

  • At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0
3 years ago
A solution was made by dissolving 5.10 mg of hemoglobin in water to give a final volume of 1.00 ml. the osmotic pressure of this
Assoli18 [71]

The molecular weight of hemoglobin can be calculated using osmotic pressure

Osmotic pressure is a colligative property and it depends on molarity as

πV = nRT

where

π = osmotic pressure

V = volume = 1mL = 0.001 L

n = moles

R = gas constant = 0.0821 L atm / mol K

T = temperature = 25°C = 25 + 273 K = 298 K

Putting values we will get value of moles

moles=\frac{\pi V}{RT}=\frac{0.00195X0.001}{0.0821X298}mol

we know that

moles=\frac{mass}{molarmass}

Therefore

molarmass=\frac{mass}{moles}=\frac{5.10X10^{-3}g}{7.97X10^{-8}}=6.399X10^{4}g

5 0
3 years ago
Help please !!!!!!!!
Elena L [17]

Answer:

Option B. 2096.1 K

Explanation:

Data obtained from the question include the following:

Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =.?

Entropy is related to enthalphy and temperature by the following equation:

Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)

ΔS = ΔH / T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

ΔS = ΔH / T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K

5 0
3 years ago
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