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soldier1979 [14.2K]
3 years ago
14

What adaptations would help an animal stay warm?

Chemistry
2 answers:
Zina [86]3 years ago
6 0

Answer:

Their adaptations help them stay warm, hide, and defend themselves. In the harsh winter months, many animals have thick coats to keep them warm.

Explanation:

LiRa [457]3 years ago
3 0
Small appendages that are close to the body stay warm and resist frostbite compared to having large ears or long tails. Thicker coats of either fur or hair grow in to act as an additional layer of insulation.
Hope this helps:)))
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Determine the percent yield forthe reaction between 82.4 g of Rband 11.6 g of O2 if 39.7 g of Rb2Ois produced
Mazyrski [523]

Step 1

The reaction is written and balanced:

4 Rb + O2 =>2 Rb2O

-----------

Step 2

Define % yield of product (Rb2O) = (Actual yield/Theoretical yield) x 100

The actual yield is provided by the exercise = 39.7 g

----------

Step 3

Determine the limiting reactant. The molar masses are needed to solve this:

For Rb) 85.4 g/mol

For O2) 32 g/mol

Procedure:

4 Rb + O2 =>2 Rb2O

4 x 85.4 g Rb ----- 32 g O2

82.4 g Rb ----- X = 7.72 g O2 are needed

For 82.4 g Rb, 7.72 g O2 is needed, but there is 11.6 g O2. Therefore, O2 is the excess agent. Rb is the limiting reactant.

--------

Step 4

Determine the theoretical yield from the limiting reactant:

The molar mass Rb2O) 187 g/mol

Procedure:

4 x 85.4 g Rb ------ 2 x 187 g Rb2O

82.4 g Rb ------ X = 90.2 g Rb2O = Theoretical yield

---------

Step 5

% yield = Actual y./Theoretical y. x 100 = (39.7 g/90.2 g) x 100 = 44 % approx.

Answer: % yield = 44 %

4 0
1 year ago
A student collects 629ml of oxygen at 0.500at, the student collected 0.0337 moles. At what temperature did the student collect t
VashaNatasha [74]

Answer:

114 K

Explanation:

Given data

  • Volume of oxygen (V): 629 mL = 0.629 L
  • Pressure of oxygen (P): 0.500 atm
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  • Temperature (T): ?

We can calculate the temperature at which the student collected the oxygen using the ideal gas equation.

P \times V = n \times R \times T\\T = \frac{P \times V}{n \times R}  = \frac{0.500atm \times 0.629L}{0.0337mol \times 0.0821atm.L/mol.K} = 114 K

The oxygen gas was collected at 114 K.

4 0
3 years ago
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Answer:

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Explanation:

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