Which types of asexual reproduction could be seen

What is the definition of an alternative acid?

Fantom [35]

alternative acids are more cost-effective and allow for flavor expression through nontraditional methods and ingredients, making for increased versatility.

6 0

3 years ago

How many atoms are in a sample of 68.7 g of copper

GarryVolchara [31]

First, you must find the number of moles of copper that your sample has. This is calculated by #of moles= your sample mass (in grams)/ atomic mass (in grams)
#of moles=68.7g/63.546g
#of moles= 1.08 moles present

Then to find the number of atoms present, we must multiply the number of moles present in the sample with Avogadro's number, which is 6.022x10^23 (this number is the number of atoms present in each mole)
# of atoms present in copper sample=1.08(6.022x10^23)
#of atoms present in copper sample= 6.50x10^23

7 0

3 years ago

A liquid mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a)

blondinia [14]

<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

We are given:

Mass percentage of $CH_4$ = 20 %

So, mole fraction of $CH_4$ = 0.2

Mass percentage of $C_2H_4$ = 30 %

So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

where,

$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$

Hence, the correct answer is Option c.

3 0

4 years ago

Which of the following claims may be made based on your observations? (You will recieve negative points for wrong answers.)

Ostrovityanka [42]

Answer:

D, E and F

Explanation:

About tetrachloro cobalt complexes, the following facts have been observed

Color of the tetrachloro cobalt complexes is blue.
They do not decompose on heating that means synthesis of tetra chloro is endothermic.

About hexa aqua cobalt complexes, the following facts have been observed

Color of the hexa aqua cobalt complexes is pink color.
They decompose on heating and remain stable on cooling that means process of synthesis of hexa aqua cobalt complexes is exothermic.

Based on above, the correct statements are:

The correct is chloro cobalt complex is blue and aqua cobalt

complex is pink.

The chloro complex is favored by heating.

If the chloro complex is a product, then the reaction must be endothermic.

The correct options are D, E and F.

8 0

3 years ago

Honors Stoichiometry Activity Worksheet

vaieri [72.5K]

Explanation: Here, we will be considering 1 cup is equal to 1 mol.

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water $=\frac{946.36 g}{236.59 g/mol}=4 mol$

Moles of sugar present in 196.86 g of water $=\frac{196.86 g}{225 g/mol}=0.8749 mol$

Moles of lemon juice present in 193.37 g of water $=\frac{193.37 g}{257.83 g/mol}=0.7499 mol$

Moles of lemonade in 2050.25 g of water $=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol$

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

$\frac{2}{1}\times 0.7499 mol = 1.4998 mol$ of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

$\frac{1}{1}\times 0.7499 mol = 0.7499 mol$ of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

$\frac{4}{1}\times 0.7499 mol=2.996 mol$ of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

$\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100$

$\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%$

The percentage yield of lemonade is 95% and ingredient which remained unused were water and sugar.

6 0

4 years ago