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Taya2010 [7]
3 years ago
7

How are estuaries connected to oceanic organisms?

Chemistry
1 answer:
Viktor [21]3 years ago
5 0

Answer:

A: Estuaries provide a nursery grounds for oceanic organisms

Explanation:

the estuaries provide Ov2 and HvO2 for these organism

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How are isotopes defined?
Alla [95]
Isotopes are atoms with the same number of protons and electrons but different number of neutrons.

Extra for better understanding:
Thus, the relative atomic mass (Ar) differs amongst the isotope. For example, Chlorine have 2 main isotopes with Ar of 35 and 37. The chemical property of the isotopes is the same but the physical property such as boiling point may change.
3 0
3 years ago
Read 2 more answers
A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what
ludmilkaskok [199]

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

  • m = mass
  • c = specific heat capacity
  • ∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

8 0
2 years ago
Which of the followings are true about D-glucose and L-glucose? A. They are furanose B. They are stereoisomers C. They are enant
zaharov [31]

Answer:

B. They are stereoisomers

C. They are enantiomers

Explanation:

Let us consider all the options

A. D and L-glucose are not necessarily furanose, they can also be in free form (open chain) or as a six-membered ring (pyranose)

B. These sugars are stereoisomers as they have the same molecular formula, same bonds but with the different spacial arrangement.

C. Two structures are called enantiomers, if they are stereoisomers and are mirror images of each other and are not-superimposable. The given pair of structures satisfy these conditions

D. Epimers are diastereoisomers (same molecular formula and connectivity having a different spacial arrangement but are not mirror images and non-superimposable) with only one different stereocenter (if there are more than one). This is not the case

E. All monosaccharides (any sugar that cannot be hydrolysed to a simpler sugar) are reducing sugars. So, this option is invalid

4 0
3 years ago
A compound composed of only carbon and chlorine is 85.5% chlorine by mass. propose a lewis structure for the lightest of the pos
aivan3 [116]

Answer is in picture below.

Use 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; mass percentage of the chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; amount of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; amount of carbon.

n(Cl) : n(C) = 2.41 mol : 1.21 mol = 2 : 1.

This compound is dichlorocarbene CCl₂.

4 0
3 years ago
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What is the half-life of an isotope that decays to 12.5% of its original activity in 19.8 hours?
katen-ka-za [31]
x= x_{0}  e^{kt} ,where~ x ~is~ the ~ |amount\ of \material\ at\ any~~time~t&#10;
and~ x_{0} ~the~original~amount.
when~x=  \frac{12.5}{100} x_{0}  }= \frac{125}{1000}  x_{0}
\frac{x}{ x_{0} } = \frac{125}{1000}= \frac{1}{8}
when t=19.8 hrs,&#10;x= x_{0}  e^{19.8 k}, &#10; \frac{x}{ x_{0} } = e^{19.8 k}, &#10; \frac{1}{8} = e^{19.8k} ,&#10;ln \frac{1}{8} =19.8 k,&#10;
ln1-ln8=19.8k,&#10;0-ln 2^{3} =19.8 k,&#10;-3 ln2=19.8k&#10;k= \frac{-3 ln2}{19.8} = \frac{- ln 2}{6.6}x= x_{0}  e^{ \frac{-ln2}{6.6}t } ,&#10; \frac{x}{ x_{0} } = e^{ \frac{- ln 2}{6.6}t } &#10;\frac{1}{2} = e^{ \frac{- ln2}{6.6}t } = \frac{1}{ e^{ \frac{ln 2}{6.6} t} } ,&#10;2= e^{ \frac{ln2}{6.6}t } ,&#10;ln2=ln  e^{ \frac{ln 2}{6.6}t } = \frac{ln 2}{6.6} t~ln e= \frac{ln2}{6.6} t,
8 0
4 years ago
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