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2 years ago

15

Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.

Record in the appropriate blanks.

2 answers:

jolli1 [7]2 years ago

8 0

Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, $0.5^{n}$

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

9966 [12]2 years ago

7 0

A=0.5
B=0.25
C=0.125
D=0.015625
E=0.00390625

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Dolomite is a mixed carbonate of calcium and magnesium. Calcium and magnesium carbonates both decompose upon heating to produce

Setler79 [48]

Answer:

72.03 %

Explanation:

Total mass of dolomite = 9.66 g

Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol$

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

$Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol$

According to the reaction shown below:-

$MgCO_3\rightarrow MgO+CO_2$

$CaCO_3\rightarrow CaO+CO_2$

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = $\frac{x}{84.3139}\ mol$

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = $\frac{x}{84.3139}\times 40.3044 \ g$

Moles of CaO = $\frac{9.66 - x}{100.0869}\ mol$

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = $\frac{9.66 - x}{100.0869}\times 56.0774 \ g$

Given that total mass of the oxide = 4.84 g

Thus,

$\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84$

$\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84$

$-694.1618435x+45673.48749\dots =40843.38968\dots$

$x=\frac{4830.09780\dots }{694.1618435}$

$x=6.9582$

Thus, the mass of Magnesium carbonate = 6.9582 g

$\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100$

$\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%$

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2 years ago

How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released durin

faust18 [17]

Answer:

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

Q = n*ΔH (2)

Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = -441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is -3,535 045.kJ/mol or -3.535x10⁶ J/mol. Solving for n we have:

n = -441,045.3 / -3.535x10⁶

n = 0.1248 moles

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RSB [31]

Answer:

$pH=11.08$

Explanation:

Hello,

In this case, since calcium hydroxide a strong base, its dissociation will completely result in both calcium and hydroxyl ions:

$Ca(OH)_2\rightarrow Ca^{2+}+OH^-$

Thus, the concentration of hydroxyl ions equals that of the calcium hydroxide, with which we could compute the pOH as shown below:

$pOH=-log([OH^-]}=-log(0.0012)\\\\pOH=2.92$

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$pOH+pH=14$

Hence, the pH finally results:

$pH=14-pOH=14-2.92\\pH=11.08$

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2 years ago

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