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3 years ago

Why do helium-filled balloons float in air?

2 answers:

8 0

A helium balloon displaces the amount of air. The weight of the helium in the balloon + the balloon fabric = lighter than the air that it displaces. Helium is lighter than air, so therefore, if it's in a balloon, the balloon will float.

kati45 [8]3 years ago

7 0

The helium balloon displaces an amount of air. As long as the weight of the helium plus the balloon fabric is lighter than the air it displaces, the balloon will float in the air. It turns out helium is a lot lighter than air.

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Hello how do I solve data for plotting graphs?

OverLord2011 [107]

You could solve it by analyzing the points by graphing them on a coordinate plane

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3 years ago

A 1.00 L flask is filled with 1.10 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres

Alex Ar [27]

Answer: a) Partial pressure of argon is 0.673 atm.

b) Partial pressure of ethane is 0.427 atm.

Explanation:

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the argon gas = ?

V= Volume of the gas = 1.00 L

T= Temperature of the gas = 25°C = 298 K (0°C = 273 K)

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{1.10g}{40g/mol}=0.0275moles$

$P=\frac{nRT}{V}=\frac{0.0275\times 0.0821\times 298}{1.00}=0.673atm$

Thus the partial pressure of argon is 0.673 atm.

b) According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$

$p_{total}=p_{Ar}+p_{ethane}$

$1.100=0.673+p_{ethane}$

$1.100-0.673=p_{ethane}$

$p_{ethane}=0.427atm$

Thus partial pressure of ethane is 0.427 atm.

7 0

3 years ago

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jeka94

The answer is Heyyy friend in know because I took the test

4 0

3 years ago

Help me Plz, I am really confused

avanturin [10]

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3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago