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3 years ago

Please help with this lab !

Chemistry

1 answer:

vekshin13 years ago

4 0

To get out the iron filings put everything in to a container and then put the magnet to the side to attract them. Pull them out of the container. Then, I don’t know how to separate the sand and salt. Maybe evaporate the salt? I joke you can figure it out. Best of luck! :)

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38- List the atomic number and name the element, whose electron configuration

strojnjashka [21]

Answer:

Atomic Number - 11. Sodium

5 0

3 years ago

What is the balanced equation of Na2S+CuSO4=CuS+Na2SO4?

lapo4ka [179]

What you were given is the balanced chemical equation

3 0

2 years ago

Atoms of iron (fe) and sulfur (s) form iron sulfide (fes) when the iron atom donates electrons to the sulfur atom. the type of b

marysya [2.9K]

Answer:

The type of bond formed in FeS would be <u>Polar Covalent Bond</u>.

Explanation:

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Covalent Bond

Between 0.4 and 1.7 then it is Polar Covalent Bond

Greater than 1.7 then it is Ionic Bond

For Fe and S,

E.N of Sulfur = 2.58

E.N of Iron = 1.83

________

E.N Difference 0.75 (Polar Covalent Bond)

4 0

3 years ago

Explain how an automobile air bag protects the people in the car from being hurt as a result of impact

liraira [26]

The quick deployment of the bag which I think is quite similar to a cushion and a balloon. It holds your head/face away from any hard surfaces. Although it does not protect your legs or other extremities that are out of the bags range.

5 0

3 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

6 0

3 years ago