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2 years ago

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A 23.8 mL sample of HCl of unknown concentration is neutralized by 5.45 mL of 0.331 M Ca(OH)2. What is the concentration of the

HCl solution?
(The answer is 0.152, but I need an explanation/steps to get that answer)

1 answer:

SCORPION-xisa [38]2 years ago

8 0

A becuase that is the one that is also b becuase a and c

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If the mass of a material is 91 grams and the volume of the material is 20 cm3, what would the density of the material be?

AlekseyPX

Use the formula Density=Mass/Volume

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3 years ago

When does phenolphthalein turn pink?

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Answer is: a) in the presence of a base.

Phenolphthalein is colorless in acidic solutions and pink in basic solutions.

Acid-base indicators are usually weak acids or bases and they are chemical detectors for hydrogen or hydronium cations.

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When solution turns phenolphthalein pink, it means it is basic (pH>7).

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2 years ago

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A student collects 350 mL of a vapor at a temperature of 67°C. The atmospheric pressure at the time of collection is 0.900 atm.

Fed [463]

Answer:

Explanation:

This problem is very similar to the other that you put before, so, we'll use the same principle here.

The ideal gas equation: PV = nRT

Where:

P: pressure in atm

V: volume in L

T: Temperature in K.

n: moles

R: Gas constant (In this case, we'll use 0.082 L atm/K)

to get the molar mass of the gas, we need to know the moles, and with the mass, we can know the molar mass. However we can put the ideal gas expression with the molar mass in this way:

we know that n is mole so:

n = g/MM

If we put this in the idea gases expression we have:

PV = gRT/MM

Solving for MM we have:

MM = gRT/PV

Now, let's convert the temperature and volume to K and L respectively:

T = 67 + 273 = 340 K

V = 350 / 1000 = 0.35 L

Now all we have to do is put all the data into the expression:

MM = 0.79 * 0.082 * 340 / 0.9 * 0.35

MM = 22.0252 / 0.315 = 69.92 g/mol rounded 70 g/mol

Now, the closest answer of your options would be 72 g/mol. This could be easily explained because we do not use all the significant figures of all numbers, including the gas constant of R. However, all the work, calculations and procedure is correct and fine, and we only have a minimum range of 2 units.

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2 years ago

Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying

PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? ( $K_a$ for HF is $6.8\times 10^{-4}$ .)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = $K_a=6.8\times 10^{-4}$

$HF\rightleftharpoons H^++F^-$

Initially

c 0 0

At equilibrium :

(c-x) x x

The expression of disassociation constant is given as:

$K_a=\frac{[H^+][F^-]}{[HF]}$

$K_a=\frac{x\times x}{(c-x)}$

$6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}$

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

$[H^+]=x=0.01346 M$

The pH of the solution is ;

$pH=-\log[H^+]=-\log[0.01346 M]=1.87$

The pH of an 0.280 M HF solution is 1.87.

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2 years ago

Use the information below to explain why the atomic radius decreases down a group.

notsponge [240]

Answer:

Detail is given below

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm

In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N

In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.

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2 years ago