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3 years ago

I need help filling out this table, please help ASAP

Chemistry

2 answers:

aleksandr82 [10.1K]3 years ago

6 0

Answer:

A-20 B-40 C-Ca D-10 E-9 F-F

Dovator [93]3 years ago

3 0

Explanation:

<h2>----------------------------------------------</h2><h2>atomic number= no.of protons</h2><h2>and </h2>

<h2>mass no.= no. of protons+no. of neutrons</h2><h2>__________________________</h2><h2><em>A</em><em>=</em><em> </em><em>2</em><em>0</em></h2><h2><em>B</em><em>=</em><em> </em><em>4</em><em>0</em></h2><h2><em>C=</em><em> </em><em>ca</em></h2><h2><em>D</em><em>=</em><em> </em><em>1</em><em>0</em></h2><h2><em>E</em><em>=</em><em> </em><em>9</em></h2><h2><em>F</em><em>=</em><em>F</em></h2>

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If a tube contains 2.0oz of sunscreen, how many kilograms of benzyl salicylate are needed to manufacture 325 tubes of sunscreen

Vadim26 [7]

<u>Answer:</u>

Now it is given here that there are 325 tubes that are manufactured for sunscreen. And we also know that the percentage of sunscreen in benzyl salicylate is 2.5% here in one tube we have 2 Oz of sunscreen.

<u>Explanation:</u>

So sunscreen in 325 tubes will be 2 x 325= 650 Oz. Since we know the amount of sunscreen in 325 tubes the amount of Benzyl salicylate in all 325 tubes will be 650 x 0.025 = 16.25 Oz

<em>Now this can be converted into kg as 0.46 kg</em>.

4 0

3 years ago

14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For

Murrr4er [49]

Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0

3 years ago

if only 0.264 g of ca(oh)2 dissolves in enough water to give 0.178 l of aqueous solution at a given temperature, what is the ksp

Maurinko [17]

The ksp value for calcium hydroxide at this temperature is 5.20 × 10⁻⁶.

Ksp is an equilibrium constant of a solid substance dissolved in a liquid solution.

Given that, the volume of water is 0.178 l, 0.264 g of Ca(OH)² dissolves in enough water

The solution equilibrium is

Ca(OH)² = Ca + 2OH

The molar solubility is

0.186 / 74.00 / 0.230 = 0.0109 M

The ksp value will be

Ksp = (s) (2S)²

Putting the values in the formula

(0.0109) x (2 x 0.0109)² = 5.20 × 10⁻⁶.

Thus, the ksp value for calcium hydroxide is 5.20 × 10⁻⁶.

To learn more about ksp value, refer to the link:

brainly.com/question/27132799

#SPJ4

8 0

1 year ago

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

5 0

3 years ago

Which of the following procedures is used during the fractional distillation of petroleum?

dem82 [27]

Increase at the temperature

7 0

3 years ago