All categories

3 years ago

How many anions are in 0.500 g of MgBr2

Chemistry

1 answer:

Oksana_A [137]3 years ago

4 0

Given:

Mass of MgBr2 = 0.500 g

To determine:

Number of anions in 0.500 g MgBr2

Explanation:

Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol

Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles

Based on stoichiometry-

1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions

Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-

Now,

1 mole of Br- contains 6.023 * 10²³ anions

0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions

Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2

You might be interested in

How to solve x² in differential

g100num [7]

Answer:

x² = mutiphy by them self

Explanation:

4 0

3 years ago

**PLATO QUESTION, PLEASE ANSWER CORRECTLY, THANKS SO MUCH**

OLga [1]

None of the questions asked can be answered completely from the graph provided (GHG emissions: Direct, indirect and total Vs Year)

Reason:
1) Question A:What caused a drop in GHG emissions around 2009?. This questions in pointing towards reason for drop of GHG emission around 2009. From the graph, it can be seen that there is a drop in GHG emission around 2009. However, information for reason for this drop is not available in graph.
2) Question B: Did GHG emissions cause the melting of Arctic glaciers?. As mentioned earlier, the graph plotted provides information of GHG emissions: Vs Year. Information related to impact of GHG on environment is not available in graph.
3) Question C: How much methane was emitted by homes between 1990 and 2000?. Graph provides information of direct and indirect emission for GHG. However, it lacks information about emission from residential or industrial sources.
4) Question D: Does industrial equipment release gases other than greenhouse gases?: Present study doesnot cover type of gases emitted from industrial equipment.
5) Question E: Which types of industries were included in the study?: Present graph has not specific information related to industries.

6 0

3 years ago

Read 2 more answers

A 25.00-mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125-mL Erlenmeyer flask and 2 drops of

Semenov [28]

Answer:

0.0611M of HNO3

Explanation:

The concentration of the NaOH solution must be 0.1198M

The reaction of NaOH with HNO3 is:

NaOH + HNO3 → NaNO3 + H2O

1 mole of NaOH reacts per mole of HNO3.

That means the moles of NaOH used in the titration are equal to moles of HNO3.

Moles HNO3:

12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.

In 25.00mL = 0.025L -The volume of the aliquot-:

0.00153 moles HNO3 / 0.025L =

7 0

3 years ago

One sound is loud and another is quieter. What must be different about these two sound waves?

tankabanditka [31]

Loudness of the sound depends on the "Amplitude" as loudness is different for two waves, their Amplitude must be different

In short, Your Answer would be "Amplitude"

Hope this helps!

3 0

2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago