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3 years ago

Study the picture of iron. How many connections, or bonds, does the center atom have to the other atoms?

Chemistry

2 answers:

Gnoma [55]3 years ago

8 0

Answer:

Explanation:

MariettaO [177]3 years ago

3 0

Answer:

Explanation:

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Which statement correctly describes diamond and graphite, which are different forms of solid carbon?

mart [117]

They differ in their molecular structures and properties.

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2 years ago

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Can someone give me an example of balancing equations with a solution that is simple?

Monica [59]

Explanation:

$H _{2}O _{2(aq)} →H _{2}O _{(l)} + O _{2}(g) \\ solution : 2 \: and\: 2$

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2 years ago

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Classify each of these reactions. A single reaction may fit more than one classification. Ba ( ClO 3 ) 2 ⟶ BaCl 2 + 3 O 2 Ba(ClO

Lena [83]

Answer: a) $Ba(ClO_3)_2\rightarrow BaCl_2+3O_2$ : Decomposition

b) $NaNO_2+HCl\rightarrow NaCl+HNO_2$ : double displacement

c) $CaO+CO_2\rightarrow CaCO_3$ : Synthesis (Combination)

d) $ZnSO_4+Mg\rightarrow MgSO_4+Zn$ : redox

Explanation:

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

$Ba(ClO_3)_2\rightarrow BaCl_2+3O_2$

A double displacement reaction is one in which exchange of ions take place.

$NaNO_2+HCl\rightarrow NaCl+HNO_2$

Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

$CaO+CO_2\rightarrow CaCO_3$

Redox reaction is a type of chemical reaction in which oxidation and reduction takes place in one single reaction. The oxidation number of one element increases and the oxidation number of other element decreases.

$ZnSO_4+Mg\rightarrow MgSO_4+Zn$

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2 years ago

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

diamong [38]

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$s}^{\mathbf{-2}}}$}$