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Kryger
2 months ago
14

Use the periodic table to identify the molar mass of each element below. Answer without doing any calculations.

Chemistry
2 answers:
Tems11 [2.7K]2 months ago
8 0

Response:

Beryllium (Be): 9.01 g/mol

Silicon (Si): 28.09 g/mol

Calcium (Ca): 40.08 g/mol

Rhodium (Rh): 102.91 g/mol

Justification:

VMariaS [2.9K]2 months ago
4 0

Response:

Beryllium (Be): 9.01 g/mol

Silicon (Si): 28.09 g/mol

Calcium (Ca): 40.08 g/mol

Rhodium (Rh): 102.91 g/mol

Justification:

You might be interested in
In the partition coefficient experiment 4A this week, after thorough mixing of the reagents, phase separation will occur. The to
VMariaS [2998]

Response:

The topmost layer is the Aqueous layer, while benzoic acid resides in the oil phase or non-aqueous layer.

Clarification:

A separating funnel is an essential tool in laboratories, utilized to split the components of immiscible liquid-liquid mixtures. This method is applied during the extraction of mixture components.

The liquids will segregate into two distinct layers. This separation occurs due to the variations in density; the heavier liquid descends to the bottom whereas the lighter liquid floats on top. The liquids involved in such separation are typically dissimilar, with one being the aqueous layer and the other the non-aqueous layer.

The partition coefficient, also known as the distribution coefficient, refers to the ratio of a compound's concentration in two immiscible solvents when in equilibrium.

Organic solvents (with the exception of halogenated organic compounds) that have densities exceeding that of water, 1g/mL (commonly referred to as the oil phase), settle beneath the aqueous layer.

Benzoic acid is found at the lower phase (i.e., the bottom layer).

7 0
2 months ago
At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a
Tems11 [2777]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2

Where P represents the pressure of each component at equilibrium.

Starting with an initial pressure of H2S at 3.00 atm, the equilibrium concentrations are:

H2S = 3.00 atm - 2X

H2 = 2X

S2 = X

Substituting these values into the equation gives:

3.2x10^{-7} = (2X)^2X / (3-2X)^2

3.2x10^{-7} = 4X^3 / 9- 6X+4X^2

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Calculating X yields:

X = 0.008945 atm

In equilibrium, the pressure of S2 is X, so the pressure stands at 0.008945 atm

7 0
1 month ago
Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce
VMariaS [2998]

Answer:

The empirical formula is = C_4H_8O

The Valproic acid formula is = C_8H_{16}O_2

Explanation:

Mass of the produced water = 0.166 g

Molar mass of water = 18 g/mol

The moles of H_2O are calculated as 0.166 g /18 g/mol = 0.00922 moles.

In 1 mole of water, there are 2 moles of hydrogen atoms.

Thus,

Moles of H = 2 x 0.00922 = 0.01844 moles

Each hydrogen atom's molar mass is 1.008 g/mol.

Hydrogen mass in the molecule = 0.01844 x 1.008 = 0.018588 g

Mass of produced carbon dioxide = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

The moles of CO_2 are calculated as 0.403 g  /44.01 g/mol = 0.009157 moles.

Each carbon atom's presence is 1 mole in 1 mole of carbon dioxide.

So,

Moles of C = 0.009157 moles

The molar mass of carbon is 12.0107 g/mol.

Carbon mass in molecule = 0.009157 x 12.0107 = 0.11 g

Since Valproic acid comprises only hydrogen, oxygen, and carbon, the oxygen mass in the sample = Total mass - Carbon mass  - Hydrogen mass.

The sample's overall mass = 0.165 g.

Oxygen mass in the sample = 0.165 - 0.11 - 0.018588 = 0.036412 g  

The molar mass of oxygen is 15.999 g/mol.

Moles of O  = 0.036412  / 15.999  = 0.002276 moles

Taking the simplest ratio for H, O, and C yields:

0.01844: 0.002276: 0.009157

= 8: 1: 4

The empirical formula becomes C_4H_8O

While molecular formulas detail the precise count of atoms for each element, empirical formulas represent the simplest form or reduced ratio of these elements in the compound.

Consequently,  

The molecular mass equals n × Empirical mass.

Here, n is a positive integer from 1, 2, 3...

Empirical mass = 4×12 + 8×1 + 16 = 72 g/mol.

Molar mass = 144 g/mol.

Thus,  

The molecular mass = n × Empirical mass.

144 = n × 72

⇒ n = 2

The Valproic acid formula is = C_8H_{16}O_2

7 0
2 months ago
Select the word or phrase from the drop-down menu to describe ionic compounds. A formula unit represents the simplest ratio of e
VMariaS [2998]

Response:

CRYSTAL

A LARGE NUMBER OF ATOMS ORGANIZED IN A REGULAR STRUCTURE

1:1

Reasoning:

8 0
2 months ago
Read 2 more answers
If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4
Anarel [2989]

Answer:

The amount of calcium sulfate that precipitates is 6.14 grams.

Explanation:

Step 1: Provided data

We are mixing 500.0 mL of 0.10 M Ca^2+ with 500.0 mL of 0.10 M SO4^2−

The Ksp for CaSO4 is 2.40×10^−5.

Step 2: Determine moles of Ca^2+

Moles of Ca^2+ = Molarity of Ca^2+ * Volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles of Ca^2+ = 0.05 moles

Step 3: Determine moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Compute total volume

Total volume = 500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Compute Q

Q = [Ca2+] [SO42-]

[Ca2+] = 0.050 M and [SO42-]

Qsp = (0.050)(0.050) = 0.0025 >> Ksp

This indicates that precipitation will take place.

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M (molar solubility)

Step 7: Determine total dissolved CaSO4

Total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 g/mol = 0.667 g

Step 8: Calculate initial mass of CaSO4

Initial moles of CaSO4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate precipitate mass

6.807 - 0.667 = 6.14 grams.

The mass of calcium sulfate that will emerge as a precipitate is 6.14 grams.

5 0
2 months ago
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