Ask question

Ask question

All categories

3 years ago

13

For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does,

will it raise or lower the actual yield:__________
Washing the crystals with hot water

1 answer:

jolli1 [7]3 years ago

5 0

Answer:

Lowers the actual yield

Explanation:

You might be interested in

Atoms of the element beryllium would most likely behave similar to the way _______ behaves.

lubasha [3.4K]

It's b. calcium
because the outer electron structure in all of that element is similar, they have the same chemical and physical properties.<span />

3 0

3 years ago

Read 2 more answers

¿Cuántos electrones conforman un enlace doble?

Aleks04 [339]

Answer:

Cada átomo aporta dos electrones al enlace, es decir, se comparten dos pares de electrones entre dos átomos. Un ejemplo es la molécula de Oxígeno (O2).

Explanation:

6 0

3 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]

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

<u>Explanation:</u>

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

<u>For calculating the mass of carbon:</u>

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.

<u>For calculating the mass of hydrogen:</u>

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:

<u>Step 1:</u> 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 = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

<u>Step 2:</u> 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 = $\frac{0.603}{0.295}=2.044\approx 2$

<u>Step 3:</u> 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

4 years ago

(14.1.50) 0.03 mol of helium are enclosed in a cylinder with a piston which maintains a constant pressure in the helium. Initial

icang [17]

Answer:

a) 0.210 j /k

b) 0.032 j/k

Explanation:

Find the attachment for solution

3 0

3 years ago

Express the pressure 545 mm Hg in kilopascals.

Arlecino [84]

545mm Hg in Kilopascals is 72.6607

I hope this helps you. Good luck stay safe, healthy and, happy!<3

6 0

2 years ago