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1 month ago

What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?

2 answers:

5 0

M = 2.76 M. To determine the molarity of any solution, the following equation is applied: M = n/V. Here, n signifies the number of moles of NH3, and V represents the solution volume in liters. Since we do not possess the moles of NH3, we must calculate it. Utilizing the mass, we can derive the moles with n = m/MM. Using the given mass and the reported molar mass of NH3 as 17.031 g/mol, we calculate: n = 2.35 g / 17.031 g/mol = 0.138 moles. With this mole quantity, the molarity calculation proceeds: M = 0.138 moles / 0.05 L, resulting in M = 2.76 mol/L. This value indicates the concentration, or molarity, of the NH₃ solution.

castortr0y [3K]1 month ago

3 0

Molarity is defined as the number of moles present in one liter of solution. Given the mass of NH₃ is 2.35 g and its molar mass is 17 g/mol, the moles of NH₃ in 2.35 g can be calculated as 2.35 g / 17 g/mol = 0.138 mol. Consequently, in a 0.05 L solution, the number of moles amounts to 0.138 mol. Therefore, the concentration in 1 L is: 0.138 mol / 0.05 L x 1L = 2.76 mol. Thus, the molarity of NH₃ is 2.76 M.

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A certain alcoholic beverage contains only ethanol (C2H6O) and water. When a sample of this beverage undergoes combustion, the e

castortr0y [3046]

Response:

9.606 g

Clarification:

Step 1: Write the balanced combustion equation

C₂H₆O(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the moles for 11.27 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

11.27 g × (1 mol/18.02 g) = 0.6254 mol

Step 3: Find the moles of C₂H₆O that produced 0.6254 moles of H₂O

The ratio of C₂H₆O to H₂O is 1:3. Thus, the moles of C₂H₆O are 1/3 × 0.6254 mol = 0.2085 mol

Step 4: Calculate the mass for 0.2085 moles of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

0.2085 mol × 46.07 g/mol = 9.606 g

7 0

3 months ago

An unknown element is found to have three naturally occurring isotopes with atomic masses of 35.9675 (0.337%), 37.9627 (0.063%)

Tems11 [2777]

Answer:

The correct choice for your inquiry is option A, Argon.

Explanation:

Isotope Atomic mass Percent (%)

1 35.9675 0.337

2 37.9627 0.063

3 39.9624 99.6

To calculate the average atomic mass: (Mass of isotope 1)(percent of 1) + (Mass of isotope 2)(percent of 2) + (Mass of isotope 3)(percent of 3)

Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) + (39.9624)(0.996)

Average atomic mass = 0.1212 + 0.0239 + 39.8025

Average atomic mass = 39.9476

Theoretical Atomic mass

a) Ar 39.95

b) K 39.10

c) Cl 35.45

d) Ca 40.08

5 0

2 months ago

A stock solution of Cu2+(aq) was prepared by placing 0.8875 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and dilut

Anarel [2989]

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

$0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}$

The molarity of Cu²⁺ is:

$\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M$

4 0

2 months ago

Magnesium metal burns with a bright white flame. What conclusions can you draw about the electron transitions that can take plac

alisha [2963]

Different wavelengths are involved.

Explanation:

When magnesium ignites with a bright white flame, it indicates that various wavelengths are related to the electron transitions occurring in the magnesium atom.

Upon combustion, the electrons within the atom become excited.
They emit characteristic light that corresponds to their energy levels.
White light consists of a mix of different wavelengths.
Seeing white light implies that multiple wavelengths combined are responsible for the observed emission.

Learn more:

Spectrum

3 0

2 months ago

Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter

KiRa [2933]

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.

8 0

2 months ago