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2 years ago

What is the halflife of a radioisotope if a 20-g sample becomes 10g after 16 hours

Chemistry

1 answer:

ohaa [14]2 years ago

5 0

Answer:

T½ = 16hours

Explanation:

Final mass (N) = 10g

Initial mass (No) = 20g

Time (t) = 16hours

T½ = ?

T½ = In2 / λ

But λ = ?

In(N/No) = -λt

In(10/20) = -(λ * 16)

In(0.5) = -16λ

-0.693 = -16λ

λ = 0.693 / 16

λ = 0.0433

Note : λ is known as the disintegration constant

T½ = In2 / λ

T½ = 0.693 / 0.0433

T½ = 16hours

The half-life of the sample is 16hours

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Which choice(s) correctly rank(s) the bonds in terms of increasing polarity?

Marina CMI [18]

Answer:

(II) only correctly rank the bonds in terms of increasing polarity.

Explanation:

Bond polarity is proportional to difference in electronegativity between bonded atoms.

Atoms Electronegativity Bond Electronegativity difference

Cl 3.0 Cl-F 1.0

Br 2.8 Br-Cl 0.2

F 4.0 Cl-Cl 0

H 2.1 H-C 0.4

C 2.5 H-N 0.9

N 3.0 H-O 1.4

O 3.5 Br-F 1.2

I 2.7 I-F 1.3

Si 1.9 Cl-F 1.0

P 2.2 Si-Cl 1.1

Si-P 0.3

Si-C 0.6

Si-F 2.1

So, clearly, order of increasing polarity : O-H > N-H > C-H

So, (II) only correctly rank the bonds in terms of increasing polarity

4 0

2 years ago

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y (1)

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol× $\frac{1mol}{80,043g}$ X

Y = 321J/g X (2)

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

X = 35,2g

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0

2 years ago

If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le

ElenaW [278]

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

$m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g$

Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

$m=\frac{92.5}{100}\times 0.5940=0.54945 g$

1 g of lead obtain from $\frac{1}{0.54945}=1.82$ grams of ore.

Thus, 2041.2 g of lead obtain from:

$2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg$

Therefore, mass of ore required to make lead sphere is 3.715 kg.

6 0

2 years ago

Morant Bay is 33 km south of Port Antonio a pigeon can fly at a speed of 1.2 m s what is the displacement of morant bay from Por

valkas [14]

Answer:

33km due south

Explanation:

Morant Bay is 33km south of Port Antonio;

Problem: displacement of Morant bay from Port Antonio;

Displacement is a vector quantity. It is the length of path between two position and the direction inclusive.

So, it has magnitude and must be specified with a direction.

Therefore, the displacement of Morant Bay from Port Antonio is 33km due south.

4 0

2 years ago

Determine the number of moles of KOH present in 95.0 ml of 0.255 M solution

gizmo_the_mogwai [7]

Answer:

Moles of KOH in 1000 mL solution = 0.255 moles

Moles of KOH in 1 mL solution = 0.255/1000 = 0.000255 moles

Moles in 95 mL solution = (95 * 255)/1000000 = 24225/1000000

Moles of KOH in 95 mL 0.255M solution = 0.024225 moles

6 0

2 years ago