What is the empirical formula for CaCo3?*

1 answer:

5 0

This problem is requiring the empirical formula for CaCO₃, which is its molecular formula, and turns out to be equal, this is A. CaCO3 according to the following:

<h3>Empirical formulas:</h3><h3 />

In chemistry, molecular formulas show both the actual type and number of atoms in a chemical compound, based on the elements across the periodic table and the subscripts standing for the number of atoms in the compound.

However, the empirical formula is a reduced expression of the molecular one, which shows the minimum number of atoms in a compound after simplifying to the smallest whole numbers.

In such a way, since the given compound is CaCO₃ and both Ca and C have a one as their subscript, it is not possible to simplify any further and therefore the empirical formula equals the molecular one this time, making the answer to be A. CaCO3.

Learn more about empirical formulas: brainly.com/question/1247523

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A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the fl

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Answer:

$\large \boxed{\text{98 m$\cdot$s}^{-2}}$

Explanation:

The formula for the velocity of the ball is

$v = \sqrt{2gh}$

1. Velocity at time of impact

$v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}$

2. Velocity on rebound

The ball has enough upward velocity to reach a height of 0.86 m.

$v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}$

3. Acceleration

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m$\cdot$s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is $\large \boxed{\textbf{98 m$\cdot$s}^{\mathbf{-2}}}$}$