Which of these is true for natural selection but not artificial selection?

Given a certain quantity of reactant, you calculate that a particular

zlopas [31]

Answer:

Percent yield = 114.5%

Explanation:

Given data:

Calculated amount of product/ theoretical yield = 55 g

Actual amount of product after performing reaction = 63 g

Percent yield = ?

Solution:

Formula:

Percent yield = ( actual yield / theoretical yield )× 100

by putting values,

Percent yield = (63 g/ 55 g)× 100

Percent yield = 114.5%

4 0

2 years ago

Consider the reaction. mc015-1.jpg How many grams of N2 are required to produce 100.0 L of NH3 at STP?

Nadusha1986 [10]

The grams of N2 that are required to produce 100.0 l of NH3 at STP

At stp 1moles = 22.4 l. what about 100.0 L of NH3

= 100 / 22.4 lx1 moles = 4.46 moles of NH3

write the reacting equation

N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2

mass = moles x molar mass

= 2.23moles x 28 g/mol = 62.4 grams

3 0

2 years ago

Choose the appropriate electron configuration for an element whose third shell contains six electrons.

slamgirl [31]

For an element whose third shell contains six electrons, the appropriate electron configuration is; 1s2 2s2 2p6 3s2 3p4.

The electron configuration shows the distribution of electrons in the shells of an atom and in orbitals.

We have been told that the six electrons are found in the third shell. This shell has n=3 and the configuration of this shell must ns2 np4.

The only electron configuration that meets this standard is 1s2 2s2 2p6 3s2 3p4.

Learn more: brainly.com/question/18704022

5 0

2 years ago

A very old tree limb contains an amount of carbon-14 that is approximately 1/8 of the current atmospheric 14C levels.. Calculate

Tomtit [17]

A. The radioactive decay equation is N = N0 $e^{-ln(2)*t/T }$

where T is the half-life (5730 years), N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewriting this as:

(N/N0) = $e^{-ln(2)*t/T }$

Since N = (1/8) N0 and substituting known values:

1/8 = $e^{-ln(2)*t/5730}$

Taking ln of both sides:

ln(1/8)= -ln(2)*t/5730

t = - 5730 * ln(1/8) / ln (2)

t = 17,190 years

The tree was cut down 17,190 years ago.

B. N0 = 1,500,000 carbon-14 atoms

Since N = (1/8) N0

N = 187,500 carbon atoms left

3 0

2 years ago

A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Cal

Nesterboy [21]

Answer:

Density is: 1.05 g/ml

Mole fraction solute: 0.015

Mole fraction solvent: 0.095

Molarity: 0.80 M

Molality: 0.82 m

Explanation:

A typical excersise of solution.

It is more confortable to make a table for this.

| masss | volume | mol

solute | | |

solvent | | |

solution | | |

Let's complete, what we have.

| masss | volume | mol

solute | 10.8g | |

solvent | | 133 mL |

solution | | 137 mL |

We can first, know how many moles are 10.8 g

Molar Mass H3PO4 = 97.99 g/mol

Mass / Molar mass = mol

10.8 g / 97.99 g/m = 0.110 mol

Density of water is 1 g/ml (it is a very knowly value)

From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

| masss | volume | mol

solute | 10.8g | | 0.110 mol

solvent | 133g | 133 mL | 7.39 mol

solution | 143.8g | 137 mL | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles or solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

5 0

2 years ago