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3 years ago

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Chemistry

1 answer:

kotegsom [21]3 years ago

5 0

Answer:

a mixture of molecules - Box f

atoms of a pure elementa metal - Box D

a solid compound - Box C

a mixture of elements - Box A

Explanation:

Box a has mixture of elements which forms a solid like shape but there are different elements present in the box. The box f has mixture of molecules in which many atoms are combined together. Box c has solid compound with single element.

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When an irregularly shaped chunk of an unknown metal with a mass of 25.32 g was placed in a graduated cylinder containing 25.00

inn [45]

Answer: The density of the unknown metal is 7.86 g/ml.

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of metal = 25.32 g

Volume of metal = volume of water displaced = (28.22 - 25.00) ml = 3.22 ml

Putting in the values we get:

$density=\frac{25.32g}{3.22ml}$

$Density=7.86g/ml$

Thus the density of the unknown metal is 7.86 g/ml

7 0

3 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

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

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago

How do we use properties of matter in our every day lives

bearhunter [10]

For example, water. In its liquid state we drink it , in its solid state (ice ) we use it to make our drinks colder or t numbs pain when wrapped in cloth and it’s gas state ( water vapour) we steam vegetables. In conclusion many different properties of matter are used for various purposes

6 0

3 years ago

Most full scuba tanks have a pressure of between 1,800 p.s.i. and 3,000 p.s.i. (pounds per square inch). Approximately 70 to 85

ArbitrLikvidat [17]

Answer: The Diver's air tank is filled with oxygen gas.

The filling of air tank while keeping in water can be explained in reference to the Charles' Law.

Charles' Law: It states that the volume of the ideal gas is directly proportional to the absolute temperature at constant pressure.

$V\propto T$ (at constant Pressure)

The Scuba tank contains cold water and is present at high pressure. The gas which is put in the tank will be compact and would not expand. As the temperature is low inside the tank, hence by using Charles' Law, the volume will also decrease inside the tank.

When the air tank is filled, it is always kept in a tub of water because the pressure is high and hence, more oxygen gas can be filled inside the air tank.

8 0

3 years ago

Read 2 more answers

Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.

leonid [27]

Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.

31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag         6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻   →   ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
       ∅        | 63.5 g Cu | 1 mol Cu | 1 mol Ag              63.5

There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.

3 0

3 years ago

Read 2 more answers