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3 years ago

how does the relationship between carbon monoxide and carbon dioxide illustrate the law of multiple proportions

1 answer:

7 0

The ratio of the weight of C to O is 1 to 1 1/3 in CO and 1 to 2 2/3 in CO2; therefore, the weights of O that will combine C are in a ratio of 1 1/3 to 2 2/3 or 1 to 2, a ratio of small whole numbers

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g Tom's roommate Bob spilled crystal violet solution on his lab coat, and Tom offers to use some lye (sodium hydroxide solution)

irina [24]

Answer:

Tom is correct. The rate of reaction of Crystal violet and NaOH is first order with respect to NaOH, hence, a higher concentration of NaOH corresponds to a higher rate of reaction; a faster reaction.

This means the speed of cleaning depends on the concentration of the lye used.

Explanation:

The reaction between Crystal violet and NaOH, represented as

CV⁺ + OH⁻ → CVOH

It is a reaction that is know to turn the violet colour of the crystal violet colourless.

The rate of the reaction is also known to be second order; first order with respect to Crystal violet and first order with respect to NaOH.

This means that the rate of reaction is directly proportional to the concentration of NaOH provided all other parameters such as the rate constant and the concentration of Crystal violet are constant.

Hence, the reaction becomes faster with an increased concentration of NaOH.

So, Tom is right, concentrated lye solution would remove the stain faster.

Bob is wrong.

Hope this Helps!!!

4 0

3 years ago

Im hvaing a hard time getting the right answer

Tanya [424]

Answer:

$V=23.9mL$

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

$n=2.49g*\frac{1mol}{125.55 g}=0.0198mol$

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

$M=\frac{n}{V}\\\\V=\frac{n}{M}$

By plugging in the moles and molarity, we obtain:

$V=\frac{0.0198mol}{0.830mol/L}=0.0239L$

Which in mL is:

$V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL$

Best regards!

6 0

3 years ago

if you mixed 72.9g hydrochloric acid(aq) with 150g silver acetate(aq), what would be the limiting reagent?

horsena [70]

Answer:

Silver Acetate would be the Limiting Reagent.

Explanation:

The balance chemical equation for the given double displacement reaction is as;

HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂

Step 1: Calculate Moles of Starting Materials:

Moles of HCl:

Moles = Mass / M.Mass

Moles = 72.9 g / 36.46

Moles = 1.99 moles

Moles of AgC₂H₃O₂:

Moles = 150 g / 166.91 g/mol

Moles = 0.898 moles

Step 2: Find out Limiting reagent as:

According to balance chemical equation.

1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂

So,

1.99 moles of HCl will react with = X moles of AgC₂H₃O₂

Solving for X,

X = 1.99 mol × 1 mol / 1 mol

X = 1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

8 0

3 years ago

A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4,

Sav [38]

Answer:

(A) 0.129 M

(B) 0.237 M

Explanation:

(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:

2HA + Ba(OH)₂ → BaA₂ + 2H₂O

Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).

We convert mass of phthalate to moles, using its molar mass:

0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol

Now we convert mmol of HA to mmol of Ba(OH)₂:

9.27 mmol HA * $\frac{1mmolBa(OH)_{2}}{2mmolHA}$ = 6.64 mmol Ba(OH)₂

Finally we calculate the molarity of the Ba(OH)₂ solution:

6.64 mmol / 35.8 mL = 0.129 M

(B) The reaction between Ba(OH)₂ and HCl is:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

So the moles of HCl that reacted are:

17.1 mL * 0.129 M * $\frac{2mmolHCl}{1mmolBa(OH)_2}$ = 4.41 mmol HCl

And the molarity of the HCl solution is:

4.41 mmol / 18.6 mL = 0.237 M

3 0

3 years ago

The length of a covalent bond depends upon the size of the atoms and the bond order.

lukranit [14]

Answer:

True

Explanation:

The length of covalent bond depends upon the size of atoms and the bond order.

6 0

3 years ago