A classmate finds that it takes 1.090 seconds for a tossed ball to touch ground. How many significant figures are in this measur

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

O sea,

Dado que:

Para :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de son:

Moles de = 0.004 moles

Para :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de son:

Moles de = 0.012 moles

Según la reacción:

1 mol de reacciona con 1 mol de

Por lo tanto,

0.012 mol de reacciona con 0.012 mol de

Moles disponibles de = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de reacciona con 1 mol de y produce 1 mol de

0.004 mol de reacciona con 0.004 mol de y genera 0.004 mol de

Los moles restantes de son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, , después de la reacción es:

11.43g/mL

Explanation:

Parameters given:

Volume of water in the graduated cylinder = 100mL

Volume of water plus lead weight = 450mL

Mass of lead weight = 4000g

Unknown:

Density of lead weight =?

Solution:

Density represents mass per unit volume of a substance.

Density  =

Volume of the lead weight equals the volume of water it displaces

 Volume of lead weight = 450 - 100 = 350mL

Density =  = 11.43g/mL

Answer:

The acetic acid concentration is measured at 8.36 M

Explanation:

Step 1: Given data

The volume of acetic acid = 1.00 mL = 0.001 L

The volume of NaOH = 32.40 mL = 0.03240 L

The molarity of NaOH = 0.258 M

Step 2: The balanced reaction equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Determining the concentration of acetic acid

b*Ca*Va = a*Cb*Vb

where b = the sodium hydroxide coefficient = 1

where Ca = the acetic acid concentration = TO BE DETERMINED

where Va = the volume of acetic acid = 1.00 mL = 0.001L

where a = the acetic acid coefficient = 1

where Cb = the sodium hydroxide molarity = 0.258 M

where Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The acetic acid concentration is 8.36 M

Sulfur in the thionyl compound carries a formal charge of +1 and possesses two lone electrons, which constitutes one lone pair.

Additional Explanation

Lewis Structures

To ascertain the formal charge and count of lone pairs, it is crucial to first construct the Lewis structure. The steps to delineate lone pairs are as follows:

1. Calculate the total valence electrons by adding the valence electrons of each atom in the compound.
2. Identify the central atom.
3. Position the remaining atoms around the central atom, connecting each with a single bond. Each bond corresponds to 2 of the total valence electrons shared.
4. Distribute any leftover valence electrons to the terminal atoms, prioritizing one until it reaches an octet, then proceed to the next. If all terminal atoms satisfy the octet rule, remaining electrons go to the central atom.

For thionyl chloride, sulfur serves as the central atom, and the overall valence electron count is:

Cl = 7 × 2 = 14

S = 6

O = 6

Total Valence Electrons = 14 + 6 + 6 = 26 electrons. Following these steps generates the corresponding Lewis structure shown in the attachment.

Having established the Lewis structure, the number of lone pairs on the central atom can be determined. In thionyl chloride, sulfur possesses two unshared electrons, signifying one lone pair.

Formal Charge Assessment

The electron distribution in a molecule is defined through the formal charges on its atoms. A formal charge reflects the disparity between the electrons surrounding an atom in a molecule versus when it is isolated. For molecular stability, formal charges across all atoms need to be minimized.

The formula for formal charge calculation is:

Formal Charge = # valence electrons for isolated atom - (# bonds + non-bonding electrons)

For the thionyl compound, sulfur's formal charge is evaluated using this formula in conjunction with the Lewis Structure:

• Valence Electrons of a Sulfur atom = 6
• Number of bonds surrounding Sulfur = 3
• Number of non-bonding electrons on Sulfur = 2

Formal Charge of S = 6 - (3 + 2)

Formal Charge of S = +1

This indicates that sulfur possesses fewer electrons in the molecule than it would as an isolated atom.

Further Exploration

1. Lewis Structure
2. Valence Electrons

Key Terms: formal charge, lone pairs

b.

c.

d.

Given lithium's atomic mass of 7 and atomic number of 3, we can express it in standard notation with the atomic mass as a superscript and the atomic number as a subscript:

a. The atomic number signifies the total number of protons present in an element. Adding a proton to lithium will increase the atomic number by one. Additionally, modifying the atomic number alters the element, since each element has a distinct atomic number.

It's critical to acknowledge that the mass of an atom is the sum of protons and neutrons; thus, we would also add one to the mass number. This gives us some new species X with:

To identify X, we must locate an element with Z = 4 on the periodic table, which is beryllium:

b. A standard lithium atom possesses an atomic number of 3, indicating it has 3 protons. Since it’s an atom, the positive charge from the protons equals the negative charge from the electrons. Thus, a neutral atom comprises the same number of protons and electrons.

Lithium begins with 3 electrons. If we remove one electron, it results in a lithium cation with a +1 charge, leading to a net charge of +3 from protons and -2 from electrons.

Hence, it can be represented as a lithium cation with a +1 charge:

c. Neutrons, while neutral, contribute to the overall mass of an atom. Therefore, adding a neutron does not affect the overall charge (atomic number) of lithium.

However, this would augment the mass by 1, as each neutron (and proton) counts as 1 atomic mass unit. Since the atomic number remains unchanged, it would still be lithium.

d. By first removing a proton, the atomic number of Li decreases by 1 unit, as the atomic number aligns with the proton count:

The mass will also reduce by 1 unit, accounting for the protons and neutrons:

Following this step, we end up with helium (which has an atomic number of 2) and a mass of 6:

Then, removing 2 neutrons will decrease the mass by 2 units while the charge of He remains unchanged, as this step does not involve protons and neutrons do not alter a nucleus's charge:

Lastly, if an electron is removed, this results in a helium cation with a +1 charge, leaving 2 protons and 1 remaining electron after removing 1 electron from helium which initially had 2 electrons in its atomic state.