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pantera1 [17]
3 years ago
6

A 6.0 M HCl solution is available. What volume is needed to make 6.00 L of 1.00 M solution?

Chemistry
1 answer:
Ket [755]3 years ago
5 0

Answer:

The volume of the stock solution needed is 1L

Explanation:

Step 1:

Data obtained from the question. This include the following:

Concentration of stock solution (C1) = 6M

Volume of stock solution needed (V1) =?

Concentration of diluted solution (C2) = 1M

Volume of diluted solution (V2) = 6L

Step 2:

Determination of the volume of the stock solution needed.

With the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:

C1V1 = C2V2

6 x V1 = 1 x 6

Divide both side by 6

V1 = 6/6

V1 = 1L

Therefore, the volume of the stock solution needed is 1L

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Becoming a physicist involves studying various subject matter, besides physics what subjects should be studied in college?​
atroni [7]

Answer:

Crystallography, nanotechnology, biophotonics, condensed matter theory and solar energy.

Explanation:

Some other subjects that should be studied in college are crystallography, nanotechnology, biophotonics, condensed matter theory ,and solar energy. These may help you in the studying of a physicist.

Hope it helped brainiest plz

3 0
3 years ago
A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make
hoa [83]

Answer:

A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make 500 mL of solution.

Calculate the pH of this buffer.

Explanation:

The pH of a buffer solution can be calculated by using the Henderson-Hesselbalch equation:

pH=pKa+log\frac{[salt]}{[acid]}

The pH of the given buffer solution can be calculated as shown below:

6 0
3 years ago
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed
KIM [24]

Answer:

On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?

A) BrCH2CH2Br

B) CH3CH2CH2Br

C) CH3CHBr2

D) CH3CH2CH2CH3

E) BrCH2CH2CH2CH2Br

Explanation:

The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.

This is an example of free radical substitution.

The structure of ethane and its bromination is shown below:

Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).

Remaining all other products are possisble to form on free radical substitution of ethane.

8 0
3 years ago
Urgent!!!
Yakvenalex [24]

The specific heat of the metal, given the data from the question is 0.60 J/gºC

<h3>Data obtained from the question </h3>

The following data were obtained from the question:

  • Mass of metal (M) = 74 g
  • Temperature of metal (T) = 94 °C
  • Mass of water (Mᵥᵥ) = 120 g
  • Temperature of water (Tᵥᵥ) = 26.5 °C
  • Equilibrium temperature (Tₑ) = 32 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Specific heat capacity of metal (C) =?

<h3>How to determine the specific heat capacity of the metal</h3>

The specific heat capacity of the sample of gold can be obtained as follow:

According to the law of conservation of energy, we have:

Heat loss = Heat gain

MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)

74 × C(94 – 32) = 120 × 4.184 (32 – 26.5)

C × 4588 = 2761.44

Divide both side by 4588

C = 2761.44 / 4588

C = 0.60 J/gºC

Thus, the specific heat capacity of the metal is 0.60 J/gºC

Learn more about heat transfer:

brainly.com/question/6363778

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6 0
1 year ago
To liquify a fixed amount of gas, one may 1. increase the temperature while decreasing the pressure. 2. increase the volume whil
antoniya [11.8K]

The complete question is:

To liquify a fixed amount of gas, one may 1. increase the temperature while decreasing the pressure. 2. increase the volume while increasing the pressure. 3. decrease the temperature while decreasing the pressure. 4. decrease the temperature while increasing the pressure. 5. decrease the volume while decreasing the pressure.

Answer:

4. decrease the temperature while increasing the pressure

Explanation:

The thermodynamics explains the sate of a substance (generally a gas or a liquid), which is its characteristics: temperature, pressure, volume, enthalpy, etc. When they changed, the state changes. Some of these changes may induce a physical change.

A gas has distant and agitated molecules, thus, the pressure is low and the temperature is higher, and a liquid has more close and less agitated molecules, so with higher pressure and low temperature.

Thus, to liquefy a fixed amount of has one may decrease the temperature while increasing the pressure.

6 0
3 years ago
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