I need help with question 39 please

A student was titrating the calcium and water solution and notice the formation of bubbles in the flask. which description would

labwork [276]

Hi!

When titrating Calcium and Water solution, if there is some CaCO3 in the solution, the following reactions may occur in acid solution:

CaCO₃ + H⁺ → Ca⁺² + HCO₃⁻

HCO₃⁻ + H⁺ ↔ H₂CO₃ → CO₂ (g) + H₂O

The bubbles are from CO₂ that is being developed from an acidic solution of CaCO₃

4 0

3 years ago

can someone answer! please don't answer for points or if you don't know it and don't answer if you are guessing make sure you un

inessss [21]

Answer:

B

Explanation:

As the distance between the planets and the sun increases, the period of revolution increases as well. The period of revolution is how long it takes for a planet to revolve around the sun. So, because the planets farther from the sun have a higher period of revolution in earth years, this also means they have longer actual years, which means the answer is B.

8 0

3 years ago

Read 2 more answers

How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?

Alex787 [66]

Answer:

about 4.8kJ

Explanation:

6 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

What was a Parthian battery?

timama [110]

Answer:

Scientists believe the batteries (if that is their correct function) were used to electroplate items such as putting a layer of one metal (gold) onto the surface of another (silver), a method still practiced in Iraq today.

Explanation:

8 0

3 years ago