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2 years ago

12

An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

ere is a 570 cm length of magnesium ribbon available?

2 answers:

melisa1 [442]2 years ago

7 0

570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students

BlackZzzverrR [31]2 years ago

5 0

In the particular experiment, each student uses 8.5 cm of ribbon.

The total length of the Mg ribbon present is 570 cm

so to find how many students can use the ribbon

number of students = total length of the ribbon / length each student uses

number of students = 570 cm / 8.5 cm/student = 67. 1 students

since it should be a whole number it should be 67

therefore 67 students can use the ribbon

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At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2

s344n2d4d5 [400]

Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
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4 0

2 years ago

A 0.25 m solution of the sugar sucrose (c12h22o11) in water is tested for conductivity using the type of apparatus shown. Bulb w

Anvisha [2.4K]

Sucrose is a non ionic compound. It does liberates ion when dissolved in water unlike NaCl or other salts which dissolve in water and produce respective cations and anions.

Thus if any amount of sucrose is dissolved in water, it will form non ionic aqueous solution (it will dissolve completely). Thus sucrose solution being non electrolytic will not conduct electricity in aqueous solution.

the bulb will not light up as sucrose will remain in molecular form only

6 0

2 years ago

Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass

inn [45]

<u>Answer:</u> The mass difference between the two is 7.38 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)

T = Temperature = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For air:</u>

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g$

Mass of air, $m_1$ = 8.56 g

<u>For helium gas:</u>

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g$

Mass of helium, $m_2$ = 1.18 g

Calculating the mass difference between the two:

$\Delta m=m_1-m_2$

$\Delta m=(8.56-1.18)g=7.38g$

Hence, the mass difference between the two is 7.38 grams.

5 0

2 years ago

One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori

Igoryamba

Answer:

$P_2=0.398bar=39800Pa$

$T_2=118.7K\\$

$Q=-3729.9J$

$W=-61753.24J$

Δ $U_T=0J$

Δ $H_T=0J$

Explanation:

Hello,

At the first state, the molar volume is:

$v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3$

The volume in both the second and third state:

$v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3$

Now, as it is about an adiabatic process, one remembers the following relationships:

$PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4$

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

$P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa$

- And the temperature:

$T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\$

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

$Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J$

Then the total heat:

$Q=Q_1+Q_2=0-3729.9J=-3729.9J$

- The work for the first process is:

$W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J$

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

$W=W_1+W_2=-61753.24J+0J=-61753.24J$

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

$U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\$

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

$H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\$

Best regards.

8 0

2 years ago

A molecule of antifreeze, ethylene glycol, has the formula c2h4(oh)2. how many atoms are there in one molecule of antifreeze?

kifflom [539]

Answer is: there are ten atoms in one molecule of antifreeze.
One molecule of ethylene glycol (C₂H₄(OH)₂) has two carbon atoms, six hydrogen atoms (4 + 2 · 1) and two oxygen atoms (2 · 1). So there are:
2 + 6 + 2 = 10 atoms.
Ethylene glycol (C₂H₄(OH)₂) is an odorless, sweet-tasting, colorless viscous liquid.

5 0

2 years ago

Read 2 more answers