Plseas help Calculate the amount of moles in 57.6 Liters of Carbon Dioxide.

45.0 g of Ca(NO3)2 are used to create a 1.3 M solution. What is the volume of the solution

Stells [14]

As we know that Molarity is given as,

M = moles / V
Solving for V,
V = moles / M ------------------(1)
Also, moles is equal to,
moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
V = (mass / M.mass) / M
Putting values,
V = (45 g / 164 g/mol) / 1.3 mol/dm³

V = 0.21 dm³

6 0

2 years ago

A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and

Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

$Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-$

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

$0.03337 M=\frac{n}{0.02328 L}$

n = $0.03337 M\times 0.02328 L=0.0007768 mol$

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

$\frac{1}{1}\times 0.0007768 mol=0.0007768 mol$ chloride ions.

In one mole of aldrin there are 6 moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

$\frac{0.0007768 mol}{6}=0.0001295 mol$

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

$\frac{0.04726 g}{0.1064 g}\times 100=44.41\%$

8 0

2 years ago

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

jolli1 [7]

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.

8 0

2 years ago

Read 2 more answers

6K + B2O3 → 3K2O + 2B

atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃ → 3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of B₂O₃ with B from balance chemical equation:

B₂O₃ : B

1 : 2

52.42 : 2×52.42 = 104.84

Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.

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