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2 years ago

6.02 x 109 = hellllppppppppppppppppppppppppppppppppppppppppppppppppppppp

Chemistry

2 answers:

worty [1.4K]2 years ago

6 0

=656.18 thts what you get if you know math hahah

mojhsa [17]2 years ago

5 0

The answer is 656.18
6.02 x 109 = 656.18

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Daniel [21]

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3 years ago

The solubility of Cr(NO3)3⋅9H2O in water is 208 g per 100 g of water at 15 ∘C. A solution of Cr(NO3)3⋅9H2O in water at 35 ∘C is

Zinaida [17]

Answer:

102g of crystals

Explanation:

When the Cr(NO₃)₃⋅9H₂O is dissolved in water at 15°C, the maximum mass that water will dissolve in the equilibrium is 208 g per 100g of water. When you heat the water, this mass will increases.

In this problem, at 35°C the water dissolves 310g in 100g of water, as in the equilibrium at 15°C the maximum mass is 208g, the mass of crystals that will form is:

310g - 208g = <em>102g of crystals</em>

<em>-Crystals are the Cr(NO₃)₃⋅9H₂O that is not dissolved-.</em>

I hope it helps!

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3 years ago

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Gnoma [55]

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2 years ago

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The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many ye

ryzh [129]

Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

$ln \frac{N}{N_0} = - k*t$ ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

$T _\frac{1}{2} = \frac{ln2}{k}$ ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of $T_\frac{1}{2}$ in above formula (equation 2) :

$28.8 yrs = \frac{ln 2}{ k }$

Multiply both side by k

$28.8 yrs * k = \frac{ln 2 }{k} * k$

Dividing both side by 28.8 yrs

$\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}$

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

$ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t$

$ln (0.0971 ) = -0.0241 yrs ^-^1 * t$

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

$\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}$

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

3 0

3 years ago

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A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water

Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be $(C_6H_{12}O)_n$

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

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2 years ago