If 33.9 ml of h3po4 neutralizes 23.4 ml of 2.28 m koh, what is the molarity of the phosphoric acid?

Chemistry

1 answer:

SpyIntel [72]2 years ago

7 0

Answer: 1.57 M

Explanation:

$M_{A}V_{A}=M_{B}V_{B}\\(33.9)M_{A}-=(23.4)(2.28)\\M_{A}=\frac{(23.4)(2.28)}{33.9}=\boxed{1.57 \text{ M}}$

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A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

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Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$

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1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

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A. the activation energy

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3 years ago

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How many mols are present in the a sample of silver nitrate which has<br> 5.3x10^24 molecules.

Westkost [7]

Answer: 5.3 x 10^24 formula units of silver nitrate is equivalent to 8.8 moles of silver nitrate. Silver nitrate is an ionic compound, therefore, its representative particle is called a "formula unit" instead of molecule. For every mole of a substance, we know that there are 6.022 x 10^23 representative units of that substance. The amount of particles in one mole of substance is called Avogadro's number.

Further Explanation:

We can convert from number of representative particles to moles using the formula:

$\boxed {no. \ of \ moles \ = \ ( given \ no. \ of \ particles) \ (\frac{1 \ mole}{\ 6.022 \ x 10^{23} particles})}$

For this problem, we can calculate the number of moles by plugging in the given values to the equation above,

$no. \ of \ moles \ = (5.3 \ x \ 10^{24} \ formula \ units \ AgNO_{3}) \ (\frac{1 \ mole \ AgNO_{3}}{6.022 \ x 10^{23} \ formula \ units AgNO_{3}}) \\\\\boxed {no. \ of moles \ AgNO_{3} \ = \ 8.8 \ moles}$