For no2−, write an equation that shows how the anion acts as a base. express your answer as a chemical equation. identify all of

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

Answer:

The correct option is C. 21900.3. I calculated 21945 J, which makes option C closely aligned with my result.

Explanation:

Data

mass = 150 g

initial temperature T1 = 10°C

final temperature T2 = 45°C

Cw = 4.18 J/g°C

Formula

Q = mCΔT = mC(T2 - T1)

Substitution

Q = (150)(4.18)(45 - 10)

Simplification

Q = (150)(4.18)(35)

Result

Q = 21945 J

Answer:

Chemists observe phenomena on a macroscopic level which informs their understanding of microscopic aspects.

Explanation:

Many critical chemical insights arise from macroscopic observations because most scientific instruments currently cannot directly evidence microscopic events. Data gathered from these larger-scale observations can yield valuable insights into the nature of specific microscopic interactions.

This is particularly true in atomic structure studies. The majority of evidence that contributed to our understanding of atomic structure was obtained from macroscopic observations and subsequently provided crucial information regarding the atom's microscopic configuration.

Explanation:

It is established that 1 gram is equivalent to 1000 milligrams. We can express this mathematically in the following way.

              or

Thus, to convert grams to milligrams, we simply multiply the number by 1000. Conversely, for converting milligrams back to grams, we divide by 1000.

<span>4.3065 g To begin with, consult the atomic masses for each involved element. Atomic weight of Calcium = 40.078 Atomic weight of Carbon = 12.0107 Atomic weight of Hydrogen = 1.00794 Atomic weight of Oxygen = 15.999 Atomic weight of Sulfur = 32.065 Next, compute the molar masses of both reactants and the product. Molar mass H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol Molar mass CaCO3 = 40.078 + 12.0107 + 3 * 15.999 = 100.0857 g/mol Molar mass CaSO4 = 40.078 + 32.065 + 4 * 15.999 = 136.139 g/mol The balanced equation for the reaction between H2SO4 and CaCO3 is: CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2 Thus, 1 mole each of CaCO3 and H2SO4 is necessary to generate 1 mole of CaSO4. Let's check the amount of moles we have for CaCO3 and H2SO4. CaCO3: 3.1660 g / 100.0857 g/mol = 0.031632891 mol H2SO4: 3.2900 g / 98.07688 g/mol = 0.033545113 mol H2SO4 is in slight excess, therefore CaCO3 is the limiting reactant, suggesting we can expect 0.031632891 moles of product. To find the mass, multiply the number of moles by the molar mass calculated previously. 0.031632891 mol * 136.139 g/mol = 4.306470148 g Given that we have 5 significant figures from our data, we round the final result to 5 figures, yielding 4.3065 g</span>