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3 years ago

9

Why can ammonium hydroxide pass through a filter but lead (II) sulfate cannot?

1 answer:

Vlad1618 [11]3 years ago

5 0

Answer:

Lead (II) sulfate is not a liquid.

Explanation:

You'd need a solubility chart for this problem, a chart that can help you determine whether or not a specific compound is a solid or not.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/Solubility_Rules

Ammonium Hydroxide is a liquid so it can pass through a filter. However, in the rules it states that sulfate salts are soluble except for the exception PbSO4. This is lead sulfate, which means that it is insoluble=solid.

Therefore lead(II) sulfate is not a liquid and cannot pass through a filter.

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Which of the following properties of matter can be explained on a submicroscopic level?

daser333 [38]

I would say all of the above

3 0

4 years ago

A sample of raw mining ore contains a hydrated salt called copper sulfate tetrahydrate, CuSO4.4H2O, along with other impurities.

julsineya [31]

Answer:

88.5 % of the raw ore is CuSO4*4H2O

Explanation:

Step 1: Data given

Mass of the ore = 10.854 grams

Mass of water = 2.994 grams

Step 2: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 2.994 grams / 18.02 g/mol

Moles H2O = 0.166 moles

Step 3: Calculate moles CuSO4*4H2O

For 1 mol CuSO4*4H2O we have 4 moles H2O

For 0.166 moles H2O we have 0.166/4 = 0.0415 moles CuSO4*4H2O

Step 4: Calculate mass CuSO4*4H2O

Mass CuSO4*4H2O = moles * molar mass

Mass CusO4*4H2O = 0.0415 moles * 231.67 g/mol

Mass CuSO4*4H2O = 9.61 grams

Step 5: Calculate mass % of CuSO4*4H2O

Mass % = (9.61 grams / 10.854 grams )*100 %

Mass % = 88.5 %

88.5 % of the raw ore is CuSO4*4H2O

4 0

4 years ago

Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years

Sliva [168]

Answer : The time passed in years is $2.83\times 10^3\text{ years}$

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = ?

a = let initial amount of the reactant = X g

a - x = amount left after decay process = $71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg$

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}$

$t=2831.00\text{ years}=2.83\times 10^3\text{ years}$

Therefore, the time passed in years is $2.83\times 10^3\text{ years}$

8 0

3 years ago

dalvyx [7]

Answer:

Mixture is the answer.

8 0

3 years ago

The acid dissociation constant, ___, is a quantitative measure of acid strength

8090 [49]

Answer:

The acid dissociation constant, _Ka__, is a quantitative measure of acid strength

Explanation:

6 0

3 years ago