Drain cleaner has a pH of about 14. How can it be classified, based on its pH?

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

(2pts) During the Purification of Lactate Dehydrogenase (LDH) experiment, you will need 50ml of buffer A150. Buffer A150 is 30mM

torisob [31]

Answer:

The answer is " $20 \ mL$ "

Explanation:

Given:

Molarity= number of moles

because it is 1 Liter

$\to \frac{0.03\ moles}{1.5 moles}=0.02\ L= 20 \ mL \ of\ Tris\\\\$

therefore,

it takes 20 mL of Tris.

$\to \frac{0.150 \ moles}{5\ moles} =0.03\ L\\\\$

$= 30 \ mL \ of\ Nacl$

So, take $20 \ mL\ of\ NaCl.$

6 0

3 years ago

What are the units used to measure specific heat capacity?

Andrews [41]

The units used to measure specific heat capacity is Joules per kilogram per Kelvin.

<h3>What is specific heat capacity?</h3>

It is the amount of heat absorbed per kilogram of material when the temperature rises by 1 Kelvin.

Specific heat capacity C is the Joules of energy in form of heat per kilogram per Kelvin temperature. The units represented by

C = ___ J/kg.K

Thus, the units used to measure specific heat capacity is Joules per kilogram per Kelvin.

Learn more about specific heat capacity.

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7 0

2 years ago

14. I have an unknown volume of gas at a pressure of 1.2 atm and a temperature of 210 K. If I raise the pressure to 1.9 atm and

Leya [2.2K]

Answer:

The law is given by the following equation: PV = nRT, where P = pressure, V = volume, n = number of moles, R is the universal gas constant, which equals 0.0821 L-atm / mole-K, and T is the temperature in Kelvin.

Explanation:

6 0

3 years ago

One beaker contains 19.93 mL, another contains 14.0 mL and a third contains 10.6 mL. What is the total volume of the three beake

Juli2301 [7.4K]

Answer:

$\large \boxed{\text{44.5 mL}}$

Explanation:

When adding or subtracting values, you must round your answer to the same "place" as the measurement with its last significant figure furthest to the left.

That is, you round off to the same number of decimal places as the measurement with the fewest decimal places.

$\begin{array}{r|r}19.9& \text{3 mL}\\14.0&\text{mL}\\10.6&\text{mL} \\\mathbf{44.5} &\textbf{3 mL}\\\end{array}$

The measurements of 14.0 and 10.6 have one digit after the decimal point, so you round the sum to have only one digit to the right of the decimal.

The number to be dropped (3) is less than 5, so you drop it.

$\text{The total volume of the three beakers is $\large \boxed{\textbf{44.5 mL}}$}$

8 0

3 years ago