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3 years ago

6

A balloon containing a sample of helium gas measures 1.50 L at a temperature of 25.0°C is placed in a refrigerator at 36.0°F. Ca

lculate its new volume. Enter your answer in the box provided.

1 answer:

Fiesta28 [93]3 years ago

6 0

Answer:

1,39 L

Explanation:

Charle's Law states that the volume of a fixed amount of gas <em>maintained at constant pressure</em> is directly proportional to the<u> absolute temperature</u> of the gas, for a constant amount of gas we can write:

$\frac{V1}{T1}=\frac{V2}{T2}$

As the pressure of the balloon doesn't change, we can use Charle's Law to solve the problem. Firs we change the given temperatures to absolute temperature units ( °K), using the following relations:

°K=273+°C

°K=5/9(°F-32)+273

Therefore:

V1=1.5 L, T1=273+25=298°K

V2=?, T2=5/9(36-32)+273=275,2°K

$V2=\frac{V1T2}{T1}=\frac{275,2*1,5}{298}=1,39L$

The new volume of the balloon is 1,39 L.

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Which kind of investigations allow for the control of variables?

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Answer:

the answer to this is O hypothesis

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2 years ago

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.

Sergio039 [100]

Explanation:

Let us take the volume of block is x.

Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.

$F_{net} = Buoyancy force(F_{b}) - weight force(w)$

Also, buoyancy force $(F_{b})$ = (volume submerged in water × density of water) + (volume in oil × density of oil)

$(F_{b})$ = $(0.592 V \times \rho) + (1 - 0.592)V \times 1000 g$

= $(0.592 V \times \rho + 408 V)$ g

As, W = V × density of graphite × g

It is given that density of graphite is $2.16 g/cm^{3}$ or 2160 $kg/m^{3}$ .

So, W = 2160 V g

$F_{net}$ = (0.592 V \rho + 408 V) g - 2160 V g = 0

$0.592 \rho$ = 1752

$\rho$ = 2959.46 $kg/m^{3}$ or 2.959 $g/cm^{3}$ is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 $cm^{3}$ oil = $35.3 \times 2.959$ 104.7 g

Hence, in second weighing total mass will be calculated as follows.

(124.8 + 104.7) g

= 229.27 g

Thus, we can conclude that in the second weighing mass is 229.27 g.

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2 years ago

How many molecules are in 2.50 moles of co2

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Answer:

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2 years ago

Read 2 more answers

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m

kogti [31]

Answer:

The molar mass of the unknown gas is $\mathbf{ 51.865 \ g/mol}$

Explanation:

Let assume that the gas is O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

$R \ \alpha \ \dfrac{1}{\sqrt{d}}$

$R = \dfrac{k}{d}$ where K = constant

If we compare the rate o diffusion of two gases;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}$

Since the density of a gas d is proportional to its relative molecular mass M. Then;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}$

Rate is the reciprocal of time ; i.e

$R = \dfrac{1}{t}$

Thus; replacing the value of R into the above previous equation;we have:

$\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}$

We can equally say:

${\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}$

${\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}$

$M_2 = 32 \times ( \dfrac{6.34}{4.98})^2$

$M_2 = 32 \times ( 1.273092369)^2$

$M_2 = 32 \times 1.62076418$

$\mathbf{M_2 = 51.865 \ g/mol}$

7 0

2 years ago

How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?

Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

$pH= -log[H]\\pH= -log (\frac{kw}{[OH]})$

$8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}$

$[OH]=1.69E-6$

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

$Cu(OH)_2 -> Cu^{2+} + 2OH^-$

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

$Kps= [Cu^{2+}] [OH]^2$

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

$Kps= s*(2s+1.69E-6)^2$

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is $s=4.96E-8 mol/L$

Now, let us calculate the moles in 1 L:

$moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles$

7 0

3 years ago