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My name is Ann [436]
3 years ago
12

All of these may be classified as

Chemistry
2 answers:
Bumek [7]3 years ago
4 0

Answer:

Where is the chart

Explanation:

mixer [17]3 years ago
3 0

What is it that you’re asking for?

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What is the mass percent of potassium sulfate in solution if 78g of potassium sulfate is dissolved in 500 mL of water? (The dens
Debora [2.8K]

Answer:

13.5 %

Explanation:

First we<u> calculate the mass of 500 mL of water</u>, using <em>its density</em>:

  • Volume * Density = Mass
  • 500 mL * 1.00 g/mL = 500 g

Then we <u>calculate the mass percent of potassium sulfate</u>, using the formula:

Mass of Potassium Sulfate / Total Mass * 100%

  • 78 g / (78 + 500) g * 100 % = 13.5 %

6 0
3 years ago
Ano-ano ang mensahe ng grupo ng el gamma penumbra sa kanilang palabas​
Alexxandr [17]

Answer:

Pag-alaga ng kalikasan, pagpapakita ng mga tanawin o kagandahan ng Pilipinas

6 0
3 years ago
PLEASE PLEASE HELP!
valentinak56 [21]

Answer: The number of grams of H_2 in 1620 mL is 1.44 g

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = 1620 ml = 1.62 L  (1L=1000ml)

n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =273K

n=\frac{PV}{RT}

n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles

Mass of hydrogen =moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g

The number of grams of H_2 in 1620 mL is 1.44 g

8 0
3 years ago
Question
hram777 [196]
I think it’s false : D
4 0
3 years ago
A sample of lead with a mass of 54.3 g is heated to a temperature of 384.4 K and placed in a container of water at 291.2 K. The
DerKrebs [107]

The mass of the water in the container given the data from the question is 22.5 g

<h3>Data obtained from the question</h3>
  • Mass of cold lead (M) = 54.3 g
  • Temperature of lead (T) = 384.4 K
  • Temperature of water (Tᵥᵥ) = 291.2 K
  • Equilibrium temperature (Tₑ) = 297.6 K
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gK
  • Specific heat capacity of lead (C) = 0.128 J/gK
  • Mass of water (Mᵥᵥ) = ?

<h3>How to determine the mass of water </h3>

Heat loss = Heat gain

MC(T – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

54.3 × 0.128 (384.4 – 297.6) = Mᵥᵥ × 4.184(297.6 – 291.2)

6.9504 × 86.8 = Mᵥᵥ × 4.184 × 6.4

Divide both side by 4.184 × 6.4

Mᵥᵥ = (6.9504 × 86.8) / (4.184 × 6.4)

Mᵥᵥ = 22.5 g

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

8 0
2 years ago
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