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Sphinxa [80]
3 years ago
11

Which of the following are products of the reaction listed? Zn + 2HCI

Chemistry
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

uh A.)? [ZnCI2+H2]? /////////

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1 point<br> How many moles of CO2 are there in 100.0 grams of CO2?
VikaD [51]

44.0095 you're welcome hope this helps

3 0
3 years ago
Read 2 more answers
If the pressure inside the cylinder increases to 1.3 atm, what is the final
EleoNora [17]

Answer:

1.4 × 10² mL

Explanation:

There is some info missing. I looked at the question online.

<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>

Step 1: Given data

  • Initial volume (V₁): 215 mL
  • Initial pressure (P₁): 625 mmHg
  • Final volume (V₂): ?
  • Final pressure (P₂): 1.3 atm

Step 2: Convert 625 mmHg to atm

We will use the conversion factor 1 atm = 760 mmHg.

625 mmHg × 1 atm/760 mmHg = 0.822 atm

Step 3: Calculate the final volume of the air

Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL

5 0
3 years ago
What is the correct term for the breakdown of organic sediment into phosphorous
viktelen [127]

Answer: Phosphorous Cycle

5 0
3 years ago
If 23.5 g of ammonia (NH3) is dissolved in 1.0 L of solution, what is the molarity?
r-ruslan [8.4K]

Answer:

1.38 M

Explanation:

Need to use the Molarity equation M=n/L

23.5g/ 17.031g/mol NH3 = 1.38 moles

1.38 moles/ 1.0 L = 1.38 M

4 0
2 years ago
What is the pH of .0003 M of NaOH
s344n2d4d5 [400]

We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH

We know that NaOH dissociates by the following reaction:

NaOH → Na⁺ + OH⁻

Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions

Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻

Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴

<u>pOH of the solution:</u>

pOH = -log[OH⁻] = -log(3 * 10⁻⁴)

pOH = -0.477 + 4

pOH = 3.523

<u>pH of the solution:</u>

We know that the sum of pH and pOH of a solution is 14

pH + pOH = 14

pH + 3.523 = 14                              [subtracting 3.523 from both sides]

pH = 10.477                        

8 0
3 years ago
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