Which statement is true about shape?

Identify each type of titration curve. note that the analyte is stated first, followed by the titrant.

Ghella [55]

Each of the type of titration curve are as follows: Strong acid-strong base= starts small than increases<span>Weak acid-strong base= low then high (small difference) Weak base- strong acid= high to low. Have in mind that titraton curves generally contain the volume of the titrant as the independent variable and the ph of the solution as the dependent variable. </span>

8 0

3 years ago

How many grams of NaCl is needed to make 750mL of a 2.4 M NaCl solution?

Aleksandr [31]

Answer:

Molarity of the solution = 3.000 M

Volume of the solution = 250.0 mL = 0.25 L

moles in 250.0 mL = molarity x volume of the solution

= 3.000 M x 0.25 L

= 0.75 mol

Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.

Moles (mol) = mass (g) / molar mass (g/mol)

Moles of NaCl in 250.0 mL = 0.75 mol

Molar mass of NaCl = 58.44 g/mol

Mass of NaCl in 250.0 mL = Moles x Molar mass

= 0.75 mol x 58.44 g/mol

= 43.83 g

Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.

Explanation:

6 0

3 years ago

Can someone help quick

alina1380 [7]

Answer:

Sodium

(Na)

Just count the electrons and search which atom it is.

7 0

3 years ago

Which of the following compounds will not exist as resonance hybrid. Give reason for your answer :

Novosadov [1.4K]

Answer:

CH3OH

Explanation:

As it lacks

$\pi$

electrons hence it will not exist as resonance hybrid.

8 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$