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3 years ago

Assuming the following materials will not mix,show how the materials would "Stack up" in a graduated cylinder.

Chemistry

1 answer:

mihalych1998 [28]3 years ago

4 0

Answer:

Density is mass per unit volume.

Its formula is

Density = Mass / Volume

Its units can be kg/L or kg/( $dm^3$ ) or g/mL or g/( $cm^3$ )

Density of various substances are given. The more denser substance goes to the bottom of the graduated cylinder.

Air on the top then comes Wood followed by Corn oil followed by Ice and then Water followed by Glycerin then Rubber then Corn syrup and at last Steel at the bottom which will more denser than all the substances mentioned above.

The staking up of the substances is in the order mentioned below in the tabular column

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Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C. None of the above pairs will

Alla [95]

Answer:

Only $Fe^{3+}+Mg$ gives spontaneous reaction.

Explanation:

A redox reaction will be spontaneous if standard reduction potential ( $E^{0}$ ) of the reaction is positive. Because it leads to negative standard gibbs free energy change ( $\Delta G^{0}$ ), which is a thermodynamic condition for spontaneity of a reaction.

$E^{0}=E^{0}(reduction)-E^{0}(oxidation)$

Where $E^{0}(reduction)$ and $E^{0}(oxidation)$ represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation: $Mg-2e^{-}\rightarrow Mg^{2+}$ ; $E_{Mg^{2+}\mid Mg}^{0}=-2.38V$

Reduction: $Fe^{3+}+3e^{-}\rightarrow Fe$ ; $E_{Fe^{3+}\mid Fe}^{0}=-0.04V$

So, $E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V$

Hence this pair will give spontaneous reaction.

(2) Similarly as above, $E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V$

Hence this pair will give non-spontaneous reaction.

(3) Similarly as above, $E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V$

Hence this pair will give non-spontaneous reaction.

(4) Similarly as above, $E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V$

Hence this pair will give non-spontaneous reaction.

6 0

2 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

A hot air balloon is filled with 15 moles helium gas and 5 moles nitrogen gas. What is the volume of the balloon at 1.01 atm and

kvasek [131]

Given:

Moles of He = 15

Moles of N2 = 5

Pressure (P) = 1.01 atm

Temperature (T) = 300 K

To determine:

The volume (V) of the balloon

Explanation:

From the ideal gas law:

PV = nRT

where P = pressure of the gas

V = volume

n = number of moles of the gas

T = temperature

R = gas constant = 0.0821 L-atm/mol-K

In this case we have:-

n(total) = 15 + 5 = 20 moles

P = 1.01 atm and T = 300K

V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L

Ans: Volume of the balloon is around 488 L

3 0

3 years ago

Write balanced reactions for the complete combustion of hydrogen. Express your answer as a chemical equation. Identify all of th

Fantom [35]

Answer:

Explanation:

A combustion involves the reaction of a fuel with oxygen (O₂). During the reaction of combustion of hydrogen (H₂), H₂ reacts with O₂ to form water (H₂O). The balanced chemical equation is the following:

2 H₂(g) + O₂(g) → 2 H₂O(g)

According to the chemical equation, 2 moles of H₂O are obtained from the reaction of 2 moles of H₂ with 1 mol of O₂. All reactants and products are in the gaseous phase.

7 0

3 years ago

What is the compound name to C5H6?

Masja [62]

What you know about rollin' down in the deep?

When your brain goes numb, you can call that mental freeze

When these people talk too much, put that stuff in slow motion, yeah

I feel like an astronaut in the ocean, ayy

3 0

2 years ago

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