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AfilCa [17]
3 years ago
8

PLS HELP THE QUESTION IS ON THE PICTURE

Chemistry
1 answer:
IceJOKER [234]3 years ago
3 0

<u>Concepts used:</u>

1 mole of an element or a compound has 6.022 * 10²³ formula units

So, we can say that: <em>Number of formula units = number of moles * 6.022*10²³</em>

number of moles of an element or a compound = given mass/molar mass

<u>__________________________________________________________</u>

<u>003 - </u><u>Number of CaH₂ formula units in 6.065 grams</u>

Number of Moles:

We know that the molar mass of CaH₂ is 42 grams/mol

Number of Moles of CaH₂ = given mass/molar mass

Number of moles = 6.065 / 42

Number of moles = 0.143 moles

Number of Formula units:

Number of formula units = number of moles * 6.022*10²³

= 0.143 * 6.022 * 10²³

= 0.86 * 10²³ formula units

__________________________________________________________

<u>004 </u><u>- Mass of 6.34 * 10²⁴ formula units of NaBF₄</u>

Number of Moles:

We mentioned this formula before:

<em>Number of formula units = number of moles * 6.022*10²³</em>

Solving it for number of moles, we get:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variable

Number of moles = 6.34 * 10²⁴ / 6.022*10²³

Number of moles=  10.5 moles

Mass of 10.5 moles of NaBF₄:

Molar mass of NaBF₄ = 38 grams/mol

Mass of 10.5 moles = 10.5 * molar mass

Mass of 10.5 moles = 10.5 * 38

Mass = 399 grams

__________________________________________________________

<u>005</u><u> - Number of moles in 9.78 * 10²¹ formula units of CeI₃</u>

Number of Moles:

We have the formula:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variables

Number of Moles = 9.78 * 10²¹ / 6.022*10²³

Number of Moles = 1.6 / 10²

Number of Moles = 1.6 * 10⁻² moles   OR   0.016 moles

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a=-1.17\times 10^{14} \hspace{3}\frac{m}{s^2}

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F_e=q*E

If the electric field is uniform, the force is constant and so is the acceleration. Applying the equations of uniformly accelerated rectilinear motion, we obtain the velocity of the particle at any time or after having moved a certain distance:

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Where:

E=Electric\hspace{3}field\hspace{3}strength=664\frac{N}{C} \\q=Electric\hspace{3}charge\hspace{3}of\hspace{3}the\hspace{3}particle\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}particle

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q=-1.6\times 10^{-19} C\\m=9.1\times 10^{-31} kg

So, replacing this data in the equation (1) :

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