A chemist adds 1.80L of a 1.1/molL aluminum chloride AlCl3 solution to a reaction flask. Calculate the millimoles of aluminum chloride the chemist has added to the flask. Round your answer to 2 significant digits.

2000 millimoles of AlCl₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 1.8 L

Molarity of solution = 1.1 mol /L

Millmole of AlCl₃ =?

Next, we shall determine the number of mole of AlCl₃ in the solution.

This can be obtained as follow:

Volume of solution = 1.8 L

Molarity of AlCl₃ solution = 1.1 mol /L

Number of mole of AlCl₃ =?

Molarity = mole /Volume

1.1 = Number of mole of AlCl₃ / 1.8

Cross multiply

Number of mole of AlCl₃ = 1.1 × 1.8

Number of mole of AlCl₃ = 1.98 moles

Finally, we shall convert 1.98 moles to millimoles. This can be obtained as follow:

1 mole = 1000 millimoles

Therefore,

1.98 mole = 1.98 mole × 1000 millimoles / 1 mole

1.98 mole = 1980 millimoles

1.98 mole ≈ 2000 millimoles

Thus, the chemist added 2000 millimoles of AlCl₃

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.