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3 years ago

10

The initial water volume in this container was 10.0 mL. When the key was submerged, 3 volumes of displaced water were collected:

0.7 mL, 0.8 mL, and 0.9 mL. What is the volume of the key?
mL

1 answer:

artcher [175]3 years ago

3 0

Answer:

2.4mL

Explanation:

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If you take Organic Chemistry Laboratory (CHM2211L) you will learn the term reflux. A reaction is considered to be at reflux whe

shutvik [7]

Answer : The temperature of the chloroform will be, $101.67^oF$

Explanation :

First we have to calculate the mass of chloroform.

$\text{Mass of chloroform}=\text{Density of chloroform}\times \text{Volume of chloroform}=1.4832g/ml\times 74.81ml=110.958g$

conversion used : $(1cm^3=1ml)$

Now we have to calculate the temperature of the chloroform.

Formula used :

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

where,

q = amount of heat or energy = 1.46 kJ = 1460 J (1 kJ = 1000 J)

$c$ = specific heat capacity = $0.96J/g.K$

m = mass of substance = 110.958 g

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25^oC=273+25=298K$

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

$1460J=110.958g\times 0.96J/g.K\times (T_{final}-298)K$

$T_{final}=311.706K$

Now we have to convert the temperature from Kelvin to Fahrenheit.

The conversion used for the temperature from Kelvin to Fahrenheit is:

$^oC=\frac{5}{9}\times (^oF-32)$

As we know that, $K=^oC+273$ or, $K-273=^oC$

$K-273=\frac{5}{9}\times (^oF-32)$

$K=\frac{5}{9}\times (^oF-32)+273$ ...........(1)

Now put the value of temperature of Kelvin in (1), we get:

$311.706K=\frac{5}{9}\times (^oF-32)+273$

$T_{final}=101.67^oF$

Therefore, the temperature of the chloroform will be, $101.67^oF$

5 0

2 years ago

Atomic weight and atomic number of an element is 35 and 17 respectively<br>

mart [117]

Chlorine.

Chlorine is the 17th element and has a mass of 35.

7 0

2 years ago

The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn

Oxana [17]

Answer:

0.193 M

Explanation:

We need to calculate the Zn²⁺ concentration at the cathode where reduction occurs which is the right side in the expression:

Zn2+(aq,0.100 M) ‖ Zn2+(aq,? M) | Zn(s)

Zn²⁺ (aq,?) + 2 e⁻ ⇒ Zn (s)

and oxidation will occur in the anode

Zn (s) ⇒ Zn²⁺ (aq, 0.100 M ) + 2 e⁻

and the overall reaction is

Zn²⁺ (aq,?) ⇒ Zn²⁺ (0.100 M )

The driving force is the difference in concentration and E the electromotive force will be given by

E = Eº - 0.0592/2 log [Zn²⁺ (0.100 M) / Zn²⁺ (M) ]

Plugging the value for E and knowing Eº is cero because we have the same electrodes, we have

17.0 x 10⁻³ = 0 - 0.0592 log 0.100/ X =

- 17.0 x 10⁻³ / 0.0592 = log 0.100 / X

- 0.287 = log (0.100 / X)

Taking inverse log to both sides of the equation

0.516 = 0.100 / X ⇒ X = 0.100 / 0.516 = 0.193 M

4 0

3 years ago

Given the following balanced equation determine the amount of MgCl2 that can be produced from 65.0 grams of Mg. Mg + HCl -->

DedPeter [7]

Answer:

Mg + HCL --> MgCl2 + H2

mass 65.0 mass 257.26

RFM 24 RFM 95

moles 2.708 moles 2.708

Explanation:

mass of MgCl = 257.26 grams

4 0

2 years ago

If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma

Veseljchak [2.6K]

Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

4 0

2 years ago