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2 years ago

What does the suffix ite stand for

Chemistry

1 answer:

insens350 [35]2 years ago

8 0

Answer: The -ite suffix is used on the oxyanion with one oxygen atom fewer (like sulfite SO32- or nitrite NO2-).

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Carbon dioxide (CO2) is a gaseous compound. Calculate the percent composition of this compound. Answer using three significant f

Tamiku [17]

Carbon dioxide is a gaseous molecule made up of the elements, C and O. Each mole of carbon dioxide has one mole C and two mole oxygen atoms.

Molar mass of carbon dioxide ( $CO_{2}$ )= $(1 * 12.01\frac{g}{mol}) +(2* 16\frac{g}{mol})=44.01\frac{g}{mol}$

Percentage by mass of carbon = $\frac{(1*12.01\frac{g}{mol}) }{44.01\frac{g}{mol} } *100=27.3%$

Percentage by mass of oxygen = $\frac{(2*16\frac{g}{mol})}{44.01\frac{g}{mol} } *100=72.7%$

Therefore C is 27.3 % and O is 72.7 % by mass in 1 mol CO $_{2}$

7 0

3 years ago

Read 2 more answers

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0

3 years ago

Be sure to answer all parts.

Llana [10]

Answer:

16 g has 2 sig figs

6701g has 4 sig figs

560 has 2 sig figs

Explanation:

4 0

2 years ago

3Fe+4H2O(yields) Fe3O4+4H2. What is the mole ratio of Fe3O4 to Fe?

WINSTONCH [101]

Answer: The correct answer is Option A.

Explanation:

Mole ratio is defined as the ratio between the stoichiometric coefficients of the molecules present in the chemical reaction.

For the given balanced chemical equation:

$3Fe+4H_2O\rightarrow Fe_3O_4+4H_2$

By Stoichiometry of the reaction:

3 moles of iron metal reacts with 4 moles of water to produce 1 mole of iron oxide and 4 moles of hydrogen gas.

The mole ratio of $Fe_3O_4:Fe=1:3$

Hence, the correct answer is Option A.

7 0

2 years ago

Read 2 more answers

When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.

EleoNora [17]

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

Answer: The % yield of the product is 87.05 %

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

We are given:

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol$

For the given chemical reaction:

$CaCO_3\rightarrow CaO+CO_2$

By stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of CaO

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of CaO

We know, molar mass of $CaO$ = 56 g/mol

Putting values in above equation, we get:

$\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

$\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%$

Hence, the % yield of the product is 87.05 %

6 0

2 years ago