11.You and a friend find a rusty wire coat hanger near the

A sample of a substance was determined and contained 1.0g of mg and 1.75g of o. what is the empirical formula? (10 pts)

Juliette [100K]

Total mass = 2.75g Mass % of Mg = 1.0g x 100/2.75g = 36.36 % Mass % of O = 1.75g x 100/2.75g = 63.64 % Mol of Mg = 36.36/24 = 1.515 Mol of O = 63.64/16 = 3.977 Ratio of Mol of Mg and O in the substance = (1.515 : 3.977) x 2 = 3 : 8 The empirical formula of substance is Mg3O8

7 0

3 years ago

A gas has a volume of 62.65L at O degrees Celsius and 1 atm. At what temperature in Celsius would the volume of the gas be 78.31

Anastaziya [24]

Answer:

The volume of the gas will be 78.31 L at 1.7 °C.

Explanation:

We can find the temperature of the gas by the ideal gas law equation:

$PV = nRT$

Where:

n: is the number of moles

V: is the volume

T: is the temperature

R: is the gas constant = 0.082 L*atm/(K*mol)

From the initial we can find the number of moles:

$n = \frac{P_{1}V_{1}}{RT_{1}} = \frac{1 atm*62.65 L}{(0.082 L*atm/K*mol)*(0 + 273)K} = 2.80 moles$

Now, we can find the temperature with the final conditions:

$T_{2} = \frac{P_{2}V_{2}}{nR} = \frac{612.0 mmHg*\frac{1 atm}{760 mmHg}*78.31 L}{2.80 moles*0.082 L*atm/(K*mol)} = 274.7 K$

The temperature in Celsius is:

$T_{2} = 274.7 - 273 = 1.7 ^{\circ} C$

Therefore, the volume of the gas will be 78.31 L at 1.7 °C.

I hope it helps you!

8 0

3 years ago

Given the following balanced equation determine the amount of MgCl2 that can be produced from 65.0 grams of Mg. Mg + HCl -->

DedPeter [7]

Answer:

Mg + HCL --> MgCl2 + H2

mass 65.0 mass 257.26

RFM 24 RFM 95

moles 2.708 moles 2.708

Explanation:

mass of MgCl = 257.26 grams

4 0

3 years ago

Chlorite anion was exposed to strong radiation and had 2 successive electrons removed, creating a chlorite cation. What is the m

katrin [286]

Answer:

bent

Explanation:

The chlorite ion has been pictured in the image attached with its two main resonance contributors.

The ion has four regions of electron density as shown. On the basis of the VSEPR theory, we expect a tetrahedral geometry.

However, the ion is bent due to the effect of the lone pairs on chlorine which is the central atom in the ion. This distorts the bond angle of the ion hence the ion has a bent molecular geometry.

7 0

3 years ago

The amount ofcalcium present in milk can be determined by adding oxalate to asample and measuring the massof calcium oxalate pre

Sauron [17]

Answer: The mass percent of calcium in milk is 0.107 %

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of calcium oxalate = 0.429 g

Molar mass of calcium oxalate = 128.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium oxalate}=\frac{0.429g}{128.1g/mol}=0.0033mol$

The given chemical equation follows:

$Na_2C_2O_4(aq.)+Ca^{2+}(aq.)\rightarrow CaC_2O_4(s)+2Na^+(aq.)$

Sodium oxalate is present in excess. So, it is considered as an excess reagent. And, calcium ion is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of calcium oxalate is produced from 1 mole of calcium ion

So, 0.0033 moles of calcium oxalate is produced from = $\frac{1}{1}\times 0.0033=0.0033mol$ of calcium ions

Now, calculating the mass of calcium ions by using equation 1, we get:

Moles of calcium ions = 0.0033 moles

Molar mass of calcium ions = 40 g/mol

Putting values in equation 1, we get:

$0.0033mol=\frac{\text{Mass of calcium ions}}{40g/mol}\\\\\text{Mass of calcium ions}=(0.0033mol\times 40g/mol)=0.132g$

To calculate the mass percentage of calcium ions in milk, we use the equation:

$\text{Mass percent of calcium ions}=\frac{\text{Mass of calcium ions}}{\text{Mass of milk}}\times 100$

Mass of milk = 125 g

Mass of calcium ions = 0.132 g

Putting values in above equation, we get:

$\text{Mass percent of calcium ions}=\frac{0.132g}{125g}\times 100=0.107\%$

Hence, the mass percent of calcium in milk is 0.107 %

7 0

3 years ago