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3 years ago

A piece of rubber has a mass of 495 kg and a volume of 355 liters. Will it float in water, which has a density of 1.0 g/mL?

1 answer:

6 0

Compare the density of the rubber to water. If it is less that 1 g/mL then it floats in the water.

You must compare density in the same units. So convert to g/mL

$( \frac{495kg}{355L} )*( \frac{1000g}{1kg} )*( \frac{1L}{1000mL} ) = 1.39 g/mL$

The rubber does not float in water.

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Which of these requires accurate coefficients in a reaction?

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Answer:

A: molar ratio

Molar ratios state the proportions of reactants and products that are used and formed in a chemical reaction.

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3 years ago

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4 years ago

Which of the following measurements contains 2 significant figures?

Trava [24]

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3 years ago

<img src="https://tex.z-dn.net/?f=H_2PO_4%5E-%28aq%29%20%5Crightarrow%20H%5E%2B%28aq%29%20%2B%20HPO_4%5E%7B2-%7D%28aq%29" id="Te

klasskru [66]

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

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3 years ago

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

$CuBr\rightleftharpoons Cu^++Br^-$