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Romashka-Z-Leto [24]
3 years ago
12

Octopus and squids breathe through

Chemistry
1 answer:
svet-max [94.6K]3 years ago
4 0

<em>Octopus and squids breathe</em> <em>like </em><em>fishes </em><em>they </em><em>breathe </em><em>from </em><em>gills </em>

<em>so </em><em>even </em><em>octopus</em><em> and</em><em> squids</em><em> </em><em>breathe </em><em>through </em><em>gills </em><em>too.</em>

<em><u>maybe </u></em><em><u>this </u></em><em><u>answer</u></em><em><u> </u></em><em><u>would</u></em><em><u> </u></em><em><u>help</u></em><em><u> </u></em><em><u>u</u></em>

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The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80
ZanzabumX [31]
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

             A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

             A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

           A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg. 
7 0
3 years ago
A 10.0 g sample of an unknown liquid is vaporized at 120.0°C and 5.0 atm. The volume of the vapour is found to be 568.0 mL. The
Minchanka [31]

Answer:

molecular formula of liquid = C₈H₁₈

Explanation:

First we determine the empirical formula of the liquid:

Number of moles of each element present in the liquid = % mass / molar mass

For Carbon, (molar mass = 12.01 g/mol) : 84.2/12.01 =7.011 moles

For Hydrogen (molar mass = 1.01 g/mol) : 15.8/1.01 = 15.643

Simplest mole ratio of the elements, C : H  is given by:

C = 7.011/7.011 = 1.0

H = 15.643/7.011 = 2.23

Multiplying through with 5, C:H = 5:11

Therefore, empirical formula is C₅H₁₁

The molecular mass of the liquid is next determined:

Using PV = nRT to find the number of moles of the liquid present

P = 5.0 atm; V = 568.0 mL = 0.568 L; R = 0.082 L*atmmol⁻¹ K⁻¹; T = 273 + 120 = 393 K

n = PV/RT = (5*0.568)/0.082*393

n = 0.088 moles

Molar mass of liquid = mass/no of moles = 10.0 g/ 0.088 moles = 113.63 gmol⁻¹

Molecular formula = n(empirical formula)

Molar mass of empirical formula, C₅H₁₁ = 71 gmol⁻¹

n = molecular mass/empirical mass = 113.63/71 = 1.6

Therefore, molecular formula =  1.6*(C₅H₁₁) = C₈H₁₈

6 0
3 years ago
How many moles of a gas would occupy 11.4 L at 273K and 2.00 atm?
bagirrra123 [75]

Answer:

1.02mol

Explanation:

Using the general gas equation below;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

P = 2.0 atm

V = 11.4L

T = 273K

n = ?

Using PV = nRT

n = PV/RT

n = 2 × 11.4/ 0.0821 × 273

n = 22.8/22.41

n = 1.017

n = 1.02mol

3 0
3 years ago
Substances that are likely to disassociate in water
mr_godi [17]

Answer:

sugar,dirt,spit

Explanation:

5 0
3 years ago
2. Using the following data, calculate the average atomic mass of magnesium (give your answer to the nearest
arlik [135]

Answer:

24.32

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 24

Abundance (A%) = 78.70%

Isotope B

Mass of B = 25

Abundance (B%) = 10.13%

Isotope C:

Mass of C = 26

Abundance (C%) = 11.17%

Average atomic mass of Mg =..?

The average atomic mass of Mg can be obtained as illustrated below:

Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

Average atomic mass = [(24 × 78.70)/100] + [(25 × 10.13)/100] + [(26 × 11.17)/100]

= 18.888 + 2.5325 + 2.9042

= 24.3247 ≈ 24.32

Therefore, the average atomic mass of magnesium (Mg) is 24.32

8 0
4 years ago
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