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3 years ago

Explain why the formation of sand is a physical change

2 answers:

6 0

Start with the definitions of Chemical Change, (A change in an objects molecular structure) and a Phase Change (Change in an objects phase of matter). The requirements for a Physical change is just any alteration done to the object that doesn’t fall into those two categories

iragen [17]3 years ago

5 0

Answer:

Sand is still sand but if you were to change the feel of sand it would not be sand

Explanation:

You might be interested in

What is the 2nd step in the process of natural selection?

Nesterboy [21]

Answer: Selection proper

Explanation:

it's an anti-chance process, but subject to many constraints

3 0

3 years ago

139%

GaryK [48]

Answer:

The percent composition of this compound is 94%

Explanation:

The reaction can be formed as

$2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}$

$\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}$

$\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}$

$\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}$

Based on no. of iron reacted,

$\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)$

n = m/M

$\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}$

% composition of $FeCl_3$

= $(9.75 / 10.39)^{*} 100$

= 94%

6 0

3 years ago

PLEASE HELP ME what would be the mass of 9.03 x 1021 molecules of hydrobromic acid (HBr)?

Nikitich [7]

We know 1 mole of any atom or molecules contains $6.033 \times 10^{23}$ atom or molecules.

1 mole of HBr i.e 81 gm/mol contains $6.033 \times 10^{23}$ atom or molecules.

So, mass of $9.03\times 10^{21}$ molecules is :

$m=\dfrac{81\times 9.03\times 10^{21} }{6.033 \times 10^{23}}\\\\m= 1.21\ gm$

Therefore, mass of $9.03\times 10^{21}$ molecules is 1.21 gm .

Hence, this is the required solution.

7 0

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:

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