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3 years ago

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Consider the following reaction. Upper C upper O subscript 2 (g) plus upper H subscript 2 (g) right arrow Upper C upper O (g) pl

us upper H subscript 2 upper O (l). What is being oxidized? carbon carbon dioxide oxygen hydrogen

2 answers:

sashaice [31]3 years ago

7 0

Answer:

hydrogen

Explanation:

alex41 [277]3 years ago

6 0

Answer:

Its (d) hydrogen

Explanation:

just took the test on edg2020

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The answer is letter C

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What effect does salt have on a marshmallow?

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2 years ago

Write a net ionic equation for the reaction between mno4- and fe2+ in acid solution

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Answer : The complete net ionic reaction between $MnO _{4} ^{-}$ and $Fe ^{2+}$ in an acid solution, can be written as :-

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3 years ago

What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm

Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u> <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

$n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}$

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = $5\times 196.96$ g

4.99 moles of gold contain = 984.8 g

Mass of ${3.01\times 10^{24}}$ atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = $19.32 g/cm^{3}$

Volume of the cuboid = $length\times breadth\times height$

Volume of the gold bar = $6.00\times 4.25\times 2.00$

Volume of the gold bar = 51 $cm^{3}$

Using formula,

$Density = \frac{mass}{Volume}$

$Mass = Density\times Volume$

$Mass = 19.32 \times 51$

Mass = 985.32 g

So, A gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u> <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

3 0

3 years ago

Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?

Rufina [12.5K]

Radioactive decay => C = Co { e ^ (- kt) |

Data:

Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

t = ln(2) / 0.44719 = 1.55 hr.

Answer: 1.55 hr

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3 years ago