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2 years ago

HOW TO DO STOICHIOMETRY

Chemistry

1 answer:

stellarik [79]2 years ago

8 0

Answer:

1.Balance the equation.

2.Convert units of a given substance to moles.

3.Using the mole ratio, calculate the moles of substance yielded by the reaction.

4.Convert moles of wanted substance to desired units.

Explanation:

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The forensic technician at a crime scene has just prepared a luminol stock solution by adding 13.0 g of luminol into a total vol

Andrews [41]

Answer:

0.978 M

Explanation:

Given data

Mass of luminol (solute): 13.0 g

Volume of the solution = volume of water: 75.0 mL = 0.0750 L

We can find the molarity of the stock solution of luminol using the following expression.

M = mass of solute / molar mass of solute × liters of solution

M = 13.0 g / 177.16 g/mol × 0.0750 L

M = 0.978 M

3 0

3 years ago

Volume of HCl used 25.0mL 4 l

borishaifa [10]

Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB = 25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

8 0

3 years ago

What are 2 of the characteristics of temporary physical change

Natali5045456 [20]

Answer:

Temporary in nature.

No new substance is formed.

Explanation:

Temporary in nature: Does not affect the internal structure of a substance, only the molecules are rearranged.

No new substance is formed: Most of the physical changes are reversible. We can obtain the substance back even after the change.

hope this helps

have an awesome day -TJ

8 0

2 years ago

Read 2 more answers

Calculate the theoretical yield for the bromination of both stilbenes

podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

4 0

2 years ago

The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co

garri49 [273]

Answer:

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater ( $m_{sw}$ ), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

$m_{sw} = \rho_{sw}\cdot V_{sw}$

Where:

$\rho_{sw}$ - Density of seawater, measured in grams per liters.

$V_{sw}$ - Volume of seawater, measured in liters.

If $V_{sw} = 1.5\times 10^{21}\,L$ and $\rho_{sw} = 1030\,\frac{g}{L}$ , then:

$m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)$

$m_{sw} = 1.545\times 10^{24}\,g$

The total mass of sodium chloride is determined by the following ratio:

$r = \frac{m_{NaCl}}{m_{sw}}$

$m_{NaCl} = r\cdot m_{sw}$

Given that $m_{sw} = 1.545\times 10^{24}\,g$ and $r = 0.035$ , the total mass of sodium chloride in all the seawater in the world is:

$m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)$

$m_{NaCl} = 5.408\times 10^{22}\,g$

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

8 0

3 years ago

HOW TO DO STOICHIOMETRY​

HOW TO DO STOICHIOMETRY