How many orbitals are in the p sublevel? a. 1 b. 3 c. 5 d. 7

A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif

noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = $[A_o]$

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = $[A]=0.06[A_o]$

Half life of the carbon-14 = $t_{1/2}=5730 years$

$k=\frac{0.693}{t_{1/2}}$

$[A]=[A_o]\times e^{-kt}$

$[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}$

$0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}$

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0

3 years ago

A silver necklace has a density of 10.5 g/mL and a volume of 1.5 mL. What is the mass of the silver necklace?

melamori03 [73]

Answer:

15.75g

Explanation:

Rearrange the formula: density = mass/volume, so that mass is the subject: mass = density × volume

Now substitute (plug in) the known values into the formula:

mass = 10.5 × 1.5 ⇒ mass = 15.75g

The mass of the silver necklace is 15.75g

Hope this helps!

4 0

2 years ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

3 years ago

Balance the following reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coeffici

arlik [135]

The given reaction is:

C4H10 + O2 → CO2 + H2O

The above equation is not balanced due to the unequal distribution of atoms on either side of equation

# atoms Reactants # atoms products

C = 4 C = 1

H = 10 H = 2

O = 2 O = 3

In order to balance it, multiply C4H10 by 2, O2 by 13, CO2 by 8 and H2O by 10 to get:

2C4H10 + 13 O2 → 8CO2 + 10H2O

7 0

3 years ago

Two standard methods are used to express the relation between protein concentration and its absorbance at 280 nm. One of these m

TEA [102]

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8 0

3 years ago