How many protons are present in an atom

The experimental procedure has you wash your thermometer and dry it after you measure the temperature of the acid and base solut

trapecia [35]

If you don't wash the thermometer, residual NaOH will react with the HCl solution. This is a highly exothermic reaction and will change the temperature of the solution, and thus throw off your measurement.

4 0

3 years ago

Calculate the density of an object with a mass of 3.8 g, that when placed in a 10.0 mL graduated cylinder with an initial volume

Arturiano [62]

Answer : The density of an object is 0.93 g/mL

Explanation : Given,

Mass of an object = 3.8 g

Initial volume = 4.5 mL

Final volume = 8.6 mL

First we have to calculate the volume of an object.

Volume of an object = Final volume - Initial volume

Volume of an object = 8.6 mL - 4.5 mL

Volume of an object = 4.1 mL

Now we have to calculate the density of an object.

Formula used:

$\text{Density}=\frac{\text{Mass of an object}}{\text{Volume of an object}}$

Now putting all the given values in this formula, we get:

$\text{Density}=\frac{3.8g}{4.1mL}$

$\text{Density}=0.93g/mL$

Therefore, the density of an object is 0.93 g/mL

3 0

3 years ago

Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4.

Shtirlitz [24]

Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

$HCOOH\leftrightarrow H^++HCOO^-$

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

$\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}$

$1.8*10^-^4=\frac{x^2}{c-x}$

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

$1.8*10^-^4=\frac{x^2}{0.12}$

Let's solve this for x. Multiply both sides by 0.12

$2.16*10^-^5=x^2$

taking square root to both sides:

$x=0.00465$

Now, we have got the concentration of $[H^+] .$

$[H^+] = 0.00465 M$

We know that, $pH=-log[H^+]$

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

7 0

3 years ago

Read 2 more answers

If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reac

mario62 [17]

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

8 0

4 years ago

Read 2 more answers

Compare to the sample of helium at STP, the same sample of helium at a higher temperature and a lower pressure.

Molodets [167]

Here we have to compare the state of helium gas at STP and high temperature and low pressure.

At STP (standard condition of temperature and pressure) i.e. 273K temperature and 1 bar pressure. At STP helium gas will behave as a real gas.

At higher temperature and low pressure Helium will behave as an ideal gas.

The ideal gas conditions are developed on taking into account two factors: (i) the gas molecules are point of mass and having no volume. (ii) there is no existence of force of attraction between the molecules.

The deviation from ideal gas to the real gas depends upon the van der waals' interaction between the gas molecules. Now in low pressure and high temperature, we can ignore the volume and also the inter-molecular force of attraction. Thus the gas sample can behaved as ideal gas.

But at elevated pressure and low temperature i.e. STP the assumptions are not valid and it will behave as real gas.

4 0

3 years ago