How many grams of the reactant water is needed to make 800.00g of calcium carbonate? The balance equation is:

The atomic number of sodium (Na) is 11. This is also the number of which of the following?

attashe74 [19]

D. Protons, Atomic number is the number of protons in an atom.

3 0

3 years ago

Look at pictures and help please

Tema [17]

Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Explanation:

More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.

Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.

Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

7 0

2 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$