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3 years ago

How would you prepare a 0.1M solution of HCl starting with a 1.0M solution? Assume you want to prepare 100 ml of solution.

1 answer:

4 0

This question is a dilution problem. In order to prepare a certain concentration and volume from a stock solution, you have to use the equation,

M1V1 = M2V2

From the equation, you need to calculate the volume you need from the stock solution to be able to have the diluted concentration then dilute it to a certain volume.

(1.0 M) (V1) = (0.1 M) (100 mL)
V1 = 0.1 mL is needed from the 1.0 M HCl solution

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If 621000 Joules of energy are added to 3.30 Liters of water at 286 Kelvin what will the final temperature of the water be? Temp

Tems11 [23]

Answer:

Explanation:To convert from cal/(g*C) to J/(kg*K), we just need to find a conversion factor for specific heat. There is really no mathematical way to do this other than to look in a physics or chemistry book and find a conversion factor. After doing this, you will see that 1 cal/(g*C) is equal to 4,186 J/(kg*K).

To find the specific heat of a material, first look at the units. There is energy per unit mass per unit temperature. So if we are given an amount of energy appllied to an object, its mass and how much the temperature of the object rises, we can calculate its specifc heat by dividing the energy by both the mass and the temperature, but don't forget to keep the units as they are:

Specific heat of the metal = (95 cal)/(10 K * 700g) = 0.014 cal/(g*K)

To find how much energy it requires to melt 250 grams of ice, we will need what is called the Latent Heat of Melting for ice. This is the amount of heat required to change unit mass of a solid into unit mass of a liquid at a constant temperature. Again, using a reference, the latent heat of melting for ice is found to be 334 kJ/kg. So the energy required to melt one kg of ice is 334 kJ. The amount of energy required to melt 0.250 kg of ice is then:

334 * 0.250 = 83.5 kJ

7 0

3 years ago

In the balanced molecular equation for the neutralization of sodium hydroxide with sulfuric acid, the products are:

earnstyle [38]

The balanced molecular equation for the neutralization of sodium hydroxide with sulfuric acid is:

Sodium hydroxide + Sulfuric acid → Sodium sulfate +water

<h3>What is the balanced molecular equation?</h3>

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products. In other words, the mass and the charge are balanced on both sides of the reaction.

In the given reaction, the reactants have been sulfuric acid and sodium hydroxide. Thus, these are written on the left side of the right arrow. The sodium sulfate and water have been the products and written on the right side of the right arrow.

The balanced molecular equation for the neutralization of sodium hydroxide with sulfuric acid is:

Sodium hydroxide + Sulfuric acid → Sodium sulfate +water

Learn more about balanced molecular equations here:

brainly.com/question/4025301

#SPJ4

7 0

2 years ago

100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.

yawa3891 [41]

Volume of H₂O added = 175 ml

<h3>Further explanation</h3>

Given

100 gm of a 55% (M/M) and 20% (M/M) nitric acid solution

Required

waters added

Solution

starting solution

mass H₂O = 45%=45 g

%mass of H₂O in new solution = 100%-20%=80%

Can be formulated for %mass H₂O :

$\tt \dfrac{45+x}{100+x}=80\%\\\\45+x=0.8(100+x)\\\\45+x=80+0.8x\\\\35=0.2x\rightarrow x=175~g$

For water mass=volume(density = 1 g/ml)

So volume added = 175 ml

7 0

2 years ago

What is the formula for P₃O₅?

marshall27 [118]

Answer:

i hope its help you

Explanation:

and goodnight,goodevening,goodmorning

6 0

2 years ago

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0

3 years ago