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3 years ago

PLEASE HELP!!’ I’ve posted this so many times and no one has responded!

Chemistry

2 answers:

sveticcg [70]3 years ago

7 0

Here I fund an article that might help.

https://pubchem.ncbi.nlm.nih.gov/compound/thiocyanate#section=Top

Diano4ka-milaya [45]3 years ago

4 0

A- FeSCN 2- becuase the thiicyante has a -1 charge and the iron (III) has a positive 3 charge

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Calculate the formula mass for the compound tin(IV) sulfate.

stellarik [79]

The molecular formula for tin(IV) sulfate is Sn(SO₄)₂.
mass of sulfur S = 32.065
as there is 2 sulfur in tin (IV) sulfate = 32.065 x 2 = 64.13
mass of oxygen = 15.999 and there are 8 oxygen atoms in the formula;
8 x 15.999 = 127.992
mass of tin = 118.71
now add all the masses = 64.13 + 127.992 + 118.71 = 310.832
formula mass of tin(IV) sulfate is 310.832

7 0

4 years ago

100.0 g water cools from 85.0°C to 20.0°C. If the specific heat of water is 4.18 J/g°C, calculate the change in energy.

Natali5045456 [20]

Answer:

-27.2 kJ

Explanation:

We can use the heat-transfer formula. Recall that:

$\displaystyle q = mC\Delta T$

Where m is the mass, C is the substance's specific heat, and ΔT is the change in temperature.

Hence substitute:

$\displaystyle \begin{aligned} q & = (100.0\text{ g})\left(\frac{4.18\text{ J}}{\text{g-$^\circ$C}}\right)(20.0\text{ $^\circ$C} - 85.0\text{ $^\circ$C}) \\ \\ & =(100.0\text{ g})\left(\frac{4.18\text{ J}}{\text{g-$^\circ$C}}\right)(-65.0\text{ $^\circ$C}) \\ \\ & = -2.72\times 10^4\text{ J} = -27.2\text{ kJ}\end{aligned}$

Therefore, the cooling of the water released about 27.2 kJ of heat.

5 0

3 years ago

Discuss availability of the lone pair of electrons to electrophile in dative bonding

Free_Kalibri [48]

Answer:

Explanation:

Here, a balance between attraction between nucleus and electrons, and electron-electron, and nuclei-nuclei repulsion play role.

All chemical bonds are formed by overlapping of orbitals. If the electronegativity of the two elements forming the bond is very different (elements from the 1st ,2nd groups with elements of 7th group) then ionic bond are formed. If the electronegativities are more similar, then overlapping is stronger, and covalent bonds are formed.

5 0

3 years ago

¿En qué ecosistema de Puerto Rico abunda más el pez Mero?

Sphinxa [80]

Answer:

Mostly abundant Coral reef ecosystem

Explanation:

We know that,

There are 7 ecosystem in puerto rico but the grouper fisher such as nassau grouper is one of them that present in coral reef ecosystem.

Nassau grouper fish is one of the grouper fish that present in puerto rico island. It is now announced as danger in the US, and as risk by the International Union for Conservation of Nature.

Hence, Coral reef ecosystem is the most abundant in Puerto Rico

3 0

4 years ago

One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0

4 years ago