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3 years ago

How many lead (Pb) atoms are in a 5.32 g sample of pure lead?

Chemistry

1 answer:

Elanso [62]3 years ago

4 0

Answer:

1.55×10²² molecules.

Explanation:

We'll begin by calculating the number of mole in 5.32 g of pure lead (Pb). This can be obtained as follow:

Mass of Pb = 5.32 g

Molar mass of Pb = 207 g/mol

Mole of Pb =?

Mole = mass /molar mass

Mole of Pb = 5.32/207

Mole of Pb = 0.0257 mole

Finally, we shall determine the number of molecules in 0.0257 mole of Pb. This can be obtained as follow:

From Avogadro's hypothesis,

I mole of Pb contains 6.02×10²³ molecules.

Therefore, 0.0257 mole will contain = 0.0257 × 6.02×10²³ = 1.55×10²² molecules.

Therefore, 5.32 g of pure lead (Pb) contains 1.55×10²² molecules.

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plz help asap (plz don't give helpless answers) and also ill give brainist to those who answer the questing right

Lera25 [3.4K]

1.) Particle B has a greater kinetic energy because when a particle is at higher temperature it contains more energy.

2.) Diagram B best shows the overflow of heat between the particles. It does because of the enclosed space that it is in and how it will circulate in it.

3.) Heat is transferred between object A and B by having object B (warmer) make a reaction with object A (colder).

5 0

3 years ago

How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?

givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

$H2SO3\leftrightarrow H^{+}+HSO3^{-}$ --------(1)

where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824

$HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}$ --------(2)

where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

$i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L$

Step 2:

For the second equivalence point setup an ICE table:

$HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}$

Initial 1.98 ? 0

Change -x -x x

Equil 1.98-x ?-x x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

$pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles$

Volume of NaOH required is:

$\frac{0.634\ moles}{0.615 moles/L}=0.389L$

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0

3 years ago

A scientist is combining two grey elements he thinks he will get a compound why is this prediction wrong

DedPeter [7]

In a compund, the proportion must be fixed, unlike mixture. Therefore, if he just combine two elements, there must be excess elements mixed in it.
Second, how did he combined two elements? In order to create a compund, either heat or electricity must be applied.
Therefore, even if he used heat / electricity, he still hasn't got the right proportion, therefore he must have mixed some excess elements into the compund that he thought he made.

4 0

3 years ago

Read 2 more answers

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid-vapor mixture at a pressure

Firdavs [7]

Answer:

(a) 0.699 kJ/K

(b) -0.671 kJ/K

Explanation:

The Refrigerant-134a flows into the evaporator as a saturated liquid-vapor mixture and flows out as a saturated vapor at a saturation pressure of 160 kPa and temperature of -15.64°C (estimated from the Saturated Refrigerant-134a Temperature Table).

(a) The entropy change of the refrigerant (ΔS $_{R-134a}$ ) = Q/T $_{1}$