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2 years ago

Calculate the gravimetric factor for Ag2O in AgS.

Chemistry

1 answer:

Lisa [10]2 years ago

7 0

The gravimetric factor for Ag2O in AgS is 0.1078

Explanation:

Solving the problem:

we need the atomic mass of the given data

Atomic mass of Ag2O = 231.735 g/mol
Atomic mass of AgS= 28 g/mol

Then add the atomic mass,

Addition=231.735+28

Addition=259.735

Then divide AgS by Ag2O (addition value)

Then division= 28/259.735

Division=0.1078

The gravimetric factor for Ag2O in AgS is 0.1078

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BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

1 year ago

The molecular mass of methyl ethanoate is 74.1 amu . calculate the molecular mass of propanoic acid, an isomer of methyl ethanoa

kobusy [5.1K]

Isomers are the compounds having same molecular formula but different structural formula.

Since, molecular formula is isomers are same, they have same mass.

Now, methyl ethanoate is an isomer of propanoic acid, hence they have same mass.

∴ Molecular mass of propanoic acid = 74.1 amu or 74.1 g/mol

4 0

2 years ago

Roundup, an herbicide manufactured by Monsanto, has the formula C3H8NO5P. How many moles of molecules are there in a 669.1-g sam

Vedmedyk [2.9K]

Answer:

2.4 ×10^24 molecules of the herbicide.

Explanation:

We must first obtain the molar mass of the compound as follows;

C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1

We know that one mole of a compound contains the Avogadro's number of molecules.

Hence;

169 g of the herbicide contains 6.02×10^23 molecules

Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.

7 0

1 year ago

Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57

vodomira [7]

Answer:

0.515 g

Explanation:

Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0

1 year ago

II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo

Leto [7]

Answer:

$Excess=3.53g$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg(s)+O_2(g) \rightarrow 2MgO(s)$

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

$n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO$

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

$m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg$

Thus, the mass in excess is:

$Excess=3.86g-0.334g\\\\Excess=3.53g$

Regards!

5 0

2 years ago