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2 years ago

1. Swimming pool manufacturers recommend maintaining the

Chemistry

1 answer:

Viefleur [7K]2 years ago

4 0

A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

"ppm" of "parts per million" is a unit of concentration equivalent to milligrams of solute per liters of solution.

A pool must maintain a chlorine concentration of 3.0 ppm (3.0 mg/L). The mass of chlorine in 3.4 × 10⁶ L is:

3.0 mg Cl₂/L × 3.4 × 10⁶ L = 1.0 × 10⁷ mg Cl₂

A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

Learn more about ppm here: brainly.com/question/13395702

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Giving brainliest and thanks to best answer <333 :>

valkas [14]

Answer:

Explanation:

radiation from the sun first warms the outer atmosphere (trophosphere)

convection(aka just heat moving through gas or liquid) brings the warmth down lower

conduction heats the ground

gl lol :))

7 0

3 years ago

Read 2 more answers

A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution

natulia [17]

Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume = 100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

7 0

3 years ago

Rozwiaz logogryf. Z wyróżnionych pól odczytaj hasło i je zapisz.

saul85 [17]

Answer:

1) Nitrogen

2) Methane

3) Dipeptide

4) Ammonia

5) Glycine

In Polish/Po polsku

1) Azot

2) metan

3) dipeptyd

4) Amoniak

5) glicyna

Explanation:

English Translation

Solve logogriffs. Read the password from the highlighted fields and write it down.

1.The chemical element being the main component of air (it occupies 78% of its volume),

2. The simplest saturated hydrocarbon,

3. Is a result of combining two amino acid molecules,

4. Gas with a pungent odor, soluble in water; hydrogen and nitrogen synthesis product.

5. Common name of aminoacetic acid

(the first has four boxes, the second has five boxes, the third has eight quarters has seven, fifths and seven).

Thank you in advance

1) The main component of air that makes up 78% of the air is Nitrogen.

2) The simplest saturated hydrocarbon is the first member of the alkane family, Methane.

3) Two amino acids combine by forming peptide bonds between the carboxyl group of one amino acid and the amino group of another through dehydration synthesis (loss of 1 molecule of water). Hence, the result of the combination of two amino acids are called Dipeptides.

4) Nitrogen and Hydrogen come together to form only one known water soluble gas with pungent smell, called Ammonia.

5) Aminoacetic acid as its name suggests is an amino acid with the acetyl group. It is the simplest amino acid. The common name for this compound is Glycine.

In Polish/Po polsku

1) Głównym składnikiem powietrza, które stanowi 78% powietrza, jest azot.

2) Najprostszym nasyconym węglowodorem jest pierwszy członek rodziny alkanów, metan.

3) Dwa aminokwasy łączą się, tworząc wiązania peptydowe między grupą karboksylową jednego aminokwasu i grupą aminową innego przez syntezę odwodnienia (utrata 1 cząsteczki wody). Stąd wynik połączenia dwóch aminokwasów nazywa się dipeptydami.

4) Azot i wodór tworzą razem jeden znany gaz rozpuszczalny w wodzie o ostrym zapachu, zwany amoniakiem.

5) Kwas aminooctowy, jak sama nazwa wskazuje, jest aminokwasem z grupą acetylową. To najprostszy aminokwas. Powszechna nazwa tego związku to glicyna.

Hope this Helps!!!

Mam nadzieję że to pomoże!!!

8 0

3 years ago

Given the equation: HCl + Na2SO4 → NaCl + H2SO4, if you start with 8 moles of hydrochloric acid, how many grams of sulfuric acid

Simora [160]

Answer:

392g sulfuric acid are produced

Explanation:

Based on the balanced equation:

2HCl + Na2SO4 → 2NaCl + H2SO4

<em>2 moles of HCl produce 1 mole of sulfuric acid</em>

<em />

To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -<em>Molar mass H2SO4 = 98g/mol- </em>we can find the mass of sulfuric acid produced:

<em>Moles sulfuric acid:</em>

8mol HCl * (1mol H2SO4 / 2mol HCl) = 4 mol H2SO4

<em>Mass sulfuric acid:</em>

4mol H2SO4 * (98g / mol) =

392g sulfuric acid are produced

4 0

3 years ago

When .080 moles of propane burn at STP, what volume of carbon dioxide is produced?

ahrayia [7]

Taking into account the reaction stoichiometry and the definition of STP, 5.4 L of carbon dioxide is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

<h3>Moles of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of C₃H₈ form 3 moles of CO₂, 0.080 moles of C₃H₈ form how many moles of CO₂?

$amount of moles of CO_{2} =\frac{0.080 moles of C_{3} H_{8}x3 moles of CO_{2} }{1 mole of C_{3} H_{8} }$

<u><em>amount of moles of CO₂= 0.24 moles</em></u>

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Volume of CO₂ produced</h3>

You can apply the following rule of three: If by definition of STP 1 mole of CO₂ occupies 22.4 L, 0.24 moles of CO₂ how much volume does it occupy?

$volume of CO_{2} =\frac{0.24 molesx22.4 L}{1 mole}$