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frez [133]
2 years ago
5

A jar is filled to the top with water, and a piece of cardboard is slid over the opening so that there is only water in the . If

the jar is now turned over, will the cardboard fall off? What will happen if there is any air in the jar?
Chemistry
2 answers:
aivan3 [116]2 years ago
7 0

Answer:

it will all fall unless these is air

Explanation:

Rom4ik [11]2 years ago
3 0

Answer:

If you do it carefully enough, a small amount of water will pour out of the glass — that's supposed to happen. But try not to let any air bubbles get into the glass. Finally, slowly remove the hand holding the cardboard in place.

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3 years ago
If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?
marysya [2.9K]
Given: 

Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL

The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:

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3 years ago
Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the
expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, Ca(OH)_2, in water is 0.185 g/100 ml. You will need to calculate the volume of 2.50\times 10^{-3}M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of Ca(OH)_2 = 0.185 g/100 mL

Volume of Ca(OH)_2 = 14.5 mL

Using unitary method:

In 100 mL, the mass of Ca(OH)_2 present is 0.185 g

So, in 14.5mL. the mass of Ca(OH)_2 present will be =\frac{0.185}{100}\times 14.5=0.0268g

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Given mass of Ca(OH)_2 = 0.0268 g

Molar mass of Ca(OH)_2 = 74 g/mol

Plugging values in equation 1:

\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol

Moles of OH^- present = (2\times 0.000362)=0.000724mol

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O

By the stoichiometry of the reaction:

Moles of OH^- = Moles of H^+ = 0.000724 mol

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = 2.50\times 10^{-3}

Putting values in equation 2, we get:

2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL

Hence, the volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

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Answer:

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