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natali 33 [55]
3 years ago
6

True or False: In every gram of sucrose (table sugar) there is 0.513 g of oxygen

Chemistry
1 answer:
Ber [7]3 years ago
4 0

Answer:

True

Explanation:

Step 1: Find molecular formula of sucrose

C₁₂H₂₂O₁₁

Step 2: Convert moles of oxygen present to grams

1 mol O = 16 g O

11 mol O = 176 g O

Step 3: Find molar mass of sucrose

C - 12.01 g/mol

H - 1.01 g/mol

O - 16.00 g/mol

12.01(12) + 22(1.01) + 11(16.00) = 342.34 g/mol C₁₂H₂₂O₁₁

Step 4: Set up dimensional analysis

1 gC_{12}H_{22}O_{11}(\frac{176 gO}{342.34gC_{12}H_{22}O_{11}} )

Step 5: Multiply/Divide and cancel out units

Grams of C₁₂H₂₂O₁₁ and grams of C₁₂H₂₂O₁₁ cancel out.

We are left with grams Oxygen

176/342.34 = 0.514109 grams Oxygen

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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
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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)

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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 :

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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

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