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3 years ago

4) I have a 52.4 g sample of lead with a volume of 10 cm3. What is its density?

Chemistry

1 answer:

ki77a [65]3 years ago

5 0

Answer:

<h2>Density = 5.24 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

<h3> $Density = \frac{mass}{volume}$ </h3>

From the question

mass = 52.4 g

volume = 10 cm³

Substitute the values into the above formula and solve for the density

That's

<h3> $Density = \frac{52.4}{10}$ </h3>

We have the final answer as

<h3>Density = 5.24 g/cm³</h3>

Hope this helps you

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Answer:

C. Valence electrons

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11. What is the maximum number of electrons that an atomic orbital can contain?

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Describe the properties of alkali metals. Based on their electronic arrangement, explain whether they exist alone in nature.

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Answer:

- They are highly reactive metals

- They have low electro negativity

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- They have low densities

Explanation:

Alkali metals are the elements in group 1 of the periodic table. They include Sodium, Lithium, Potassium e.t.c.

Due to the fact they have one atom in their outermost shell, they are very unstable because they easily react with other elements and are therefore don't exist alone in nature but combined with other elements for this same reason.

Since alkali metals don't easily attract other elements due to it's lone pair in the outer most shell, it can be said to have low electro negativity.

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2 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

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3 years ago

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55. Which Element has a larger radius- Rb or I?

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Rb has larger radius.

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