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3 years ago

Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure

Chemistry

1 answer:

o-na [289]3 years ago

6 0

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

(P1 × V1) / T1 = (P2 × V2) / T2

solve for V2 by making it the subject of formula,

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

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You decide to clean the bathroom. You notice that the shower is covered in a strange green slime . you try to get rid of this sl

Free_Kalibri [48]

The independent variable (IV) is the lemon juice mixture
The dependent variable (DV) is the appearance of the green slime on the shower
The control variable (CV) are time taken to spray, the amount of spray
The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
The control group (CG) is the side of the shower sprayed with water.

INDEPENDENT VARIABLE

Independent variable is the variable of an experiment that is changed by the experimenter in order to bring about a change. It is the variable being tested in the experiment. In this case, the IV is the lemon juice mixture tested on the green slime on the shower.

DEPENDENT VARIABLE:

Dependent variable is the variable that is observed or measured in an experiment. It is also called responding variable. The DV in this case is the appearance of the green slime on the shower.

CONTROL VARIABLE:

Control variable is the variable that is kept constant throughout the experiment for all groups. The CV is the same for all the groups and they include: time taken to spray, the same amount of spray

CONTROL GROUP

Control group is the group that does not receive the independent variable or test in an experiment. In this case, the CG is the side of the shower sprayed with water.

EXPERIMENTAL GROUP:

Experimental group is the group of ab experiment that receives the experimental treatment or independent variable. In this case, the EG is the side of the shower sprayed with lemon juice mixture.

Therefore, the IV, DV, CV, EG and CG of this experiment are as follows:

The independent variable (IV) is the lemon juice mixture
The dependent variable (DV) is the appearance of the green slime on the shower
The control variable (CV) are time taken to spray, the amount of spray
The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
The control group (CG) is the side of the shower sprayed with water.

Learn more: brainly.com/question/17498238?referrer=searchResults

7 0

2 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o

HACTEHA [7]

Answer:

1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Explanation:

From the question given above, the following were obtained:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Volume of solution = 20.4 mL

Concentration of solution =?

Next, we shall determine the mass of sodium hydroxide, NaOH. This can be obtained as as illustrated below:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Mass of NaOH =?

Mass of NaOH = (Mass Cylinder + NaOH) – (Mass of empty cylinder)

Mass of NaOH = 31.92 – 9.5

Mass of NaOH = 22.42 g

Finally, we shall determine concentration of the solution as follow:

Mass of NaOH = 22.42 g

Volume of solution = 20.4 mL

Concentration of solution =?

Concentration = mass /volume

Concentration of solution = 22.42 / 20.4

Concentration of solution = 1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Therefore, the concentration of the solution is 1.1 gmL¯¹

3 0

3 years ago

The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is __________

PIT_PIT [208]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl

Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality = 0.7543 mol / kg

The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is 0.7453 molal.

6 0

2 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

Element, Compound or Molecule Identify each of the images as an element, molecule or compound.

Olin [163]

Answer:

The answer is:

A) Element

B) Molecule

C) Compound

5 0

2 years ago