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Lesechka [4]
3 years ago
13

What type of consumer are bears?

Chemistry
1 answer:
sashaice [31]3 years ago
5 0
Bears are omnivores (and scavengers), meaning they eat both plants and animals.
it can be both a primary consumer (if it eats plants) and secondary & tertiary consumers (if it eats a plant-eating animal). so it depends on what it eats.
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A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)
son4ous [18]

Answer:

\large \boxed{\text{4.5 atm}}

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?;          T₂ =  20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ =  (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}

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3 years ago
What is the total number of atoms in a 75 g sample of zinc?
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3 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol
AnnyKZ [126]
<span>1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol
The independent variable would be ethylene glycol and dependent variable would be density.

A. A 25-mL volumetric flask with its stopper has a mass of 32.6341 g. The same flask filled to the line with ethylene glycol (C2H6O2, automotive antifreeze) solution has a mass of 58.0091 g. What is the density of the ethylene glycol solution?

Density = 58.0091 - 32.6341 / .025 = 1015 g/L

B. What is the molarity of the ethylene glycol solution, if the mass of ethylene glycol in the solution is 12.0439 g?

Molarity = 12.0439 ( 1 mol / 62.07 g) / 0.025 = 7.8 M</span>
5 0
3 years ago
Read 2 more answers
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