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1 year ago

The beaker shown below contains a 0.58 M solution of dye in water. How many moles of dye are there in the beaker?

2 answers:

7 0

According to the equation of molarity:

Molarity= no.of moles / volume per liter of Solution

when we have the molarity=0.58 M and the beaker at 150mL so V (per liter) = 150mL/1000 = 0.150 L

by substitution:
∴ No.of moles = Molarity * Volume of solution (per liter)
= 0.58 * 0.150 = 0.087 Moles

Sergio039 [100]1 year ago

4 0

Molarity is the number of moles that are contained in 1000 cm³ or a liter of water.
In this case the molarity is 0.58 M meaning 0.58 moles of the dye are contained in 1000 cm³ or a liter of water.
The beaker is at 150 cm³, there the number of moles will be;
=0.58 × 150/1000
= 0.087 moles

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93.2 mL of a 2.03 M potassium fluoride (KF) solution

Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

Initial volume (V₁): 93.2 mL

Initial concentration (C₁): 2.03 M

Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

$93.2mL \times \frac{1L}{1000mL} = 0.0932 L$

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

$V_2 = 0.0932L + 3.92 L = 4.01L$

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

$C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M$

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2 years ago

Read 2 more answers

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.

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2 years ago

A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

zhannawk [14.2K]

Answer: The molecular formula will be $C_8H_8O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

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2 years ago

A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration

Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= $100$ ml

Initial moles of NO= $\frac{2.60}{10^2}$ moles

Initial moles of H2= $\frac{1.30}{10^2}$ moles

Concentration of NO at equilibrium= $0.161$ M

$Concentration(in M)=\frac{moles}{volume(in litre)}$

Moles of NO at equilibrium= $0.161(\frac{100}{1000})$

= $\frac{1.61}{10^2}$ moles

2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)

Initial :1.3*10^-2 2.6*10^-2 0 0 moles

Equilibrium:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles

∴ $\frac{2.60}{10^2}-x=\frac{1.61}{10^2}$

⇒ $x=\frac{0.99}{10^2}$

$Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)$

Equilibrium:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles

⇒ $Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2} (0.1)$

⇒ $Kc=19.4760$

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2 years ago

Read 2 more answers

﻿The beaker shown below contains a 0.58 M solution of dye in water. How many moles of dye are there in the beaker?

The beaker shown below contains a 0.58 M solution of dye in water. How many moles of dye are there in the beaker?