All categories

3 years ago

Calculate the mass (g) of 1.5 moles of Sulfur.

Chemistry

1 answer:

just olya [345]3 years ago

7 0

Answer:

48 g S

Explanation:

Step 1: Define

Molar Mass of Sulfur (S) - 32.07 g/mol

Step 2: Use Dimensional Analysis

$1.5 \hspace{3} mol \hspace{3} S(\frac{32.07 \hspace{3} g \hspace{3} S}{1 \hspace{3} mol \hspace{3} S} )$ = 48.105 g S

Step 3: Simplify

We have 2 sig figs.

48.105 g S ≈ 48 g S

You might be interested in

Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04

evablogger [386]

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

8 0

3 years ago

Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

3 years ago

Give the product expected when the following alcohol reacts with pyridinium chlorochromate (PCC). (Assume that PCC is present in

stiks02 [169]

Answer:

Explanation:

From the information given, we are to show the product of the reaction between the alcohol (stick formula) with pyridinium chlorochromate. The reaction starts with the nucleophilic addition of Cr with oxygen molecule on the alcohol. Then, deprotonation of oxygen occurs, this process is followed by the removal of hydrogen ion to yield the final product. The reaction mechanism can be seen from the attached image below.

4 0

3 years ago

Aleksandr-060686 [28]

B. a chemical change

7 0

3 years ago

If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your

Drupady [299]

N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

mol N₂(MW=28 g/mol) :

$\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571$

mol H₂(MW= 2 g/mol) :

$\tt mol=\dfrac{100}{2}=50$

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

$\tt \dfrac{3.571}{1}\div \dfrac{50}{3}=3.571\div 16.6$

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant

5 0

3 years ago

Calculate the mass (g) of 1.5 moles of Sulfur. ​

Calculate the mass (g) of 1.5 moles of Sulfur.