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Pani-rosa [81]
3 years ago
9

The material used as a biological sensor is one dimension of 20 mm, the other Are in nanometers ​

Chemistry
1 answer:
mario62 [17]3 years ago
4 0

Answer: 20 , 40

I want know the answer now !

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Hg-178 undergoes 2 alpha decay and positron emission. Determine the new isotope.
Illusion [34]
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3 years ago
Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.
Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%

\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%

\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

3 0
3 years ago
Model B: In this model the atom is a solid sphere that cannot be divided up into smaller
vivado [14]

Answer:

They are called corpuscles

Explanation:

5 0
4 years ago
he mineral rhodochrosite [manganese(II) carbonate, MnCO3] is a commercially important source of manganese. Write a half-reaction
valina [46]

Answer:

MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

Explanation:

Half reaction:MnCO_{3}\rightarrow MnO_{2}

(1)CO_{3} balance: MnCO_{3}\rightarrow MnO_{2}+HCO_{3}^{-}

(2)H and O balance in acidic medium:MnCO_{3}+2H_{2}O\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

(3) charge balance:MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

Hence balanced half-reaction:MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

7 0
3 years ago
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6CO2 + 6H2O → C6H12O6 + 6O2

glucose produced = 1/6 x 23.6 = 3.93 moles

4 0
2 years ago
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