What should be the labels on the X-axis and Y-axis?

If 17. 6 g of hcl are used to produce a 14. 5 l solution, what is the ph of the solution?.

kati45 [8]

This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.

<h3>pH calculations</h3>

In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:

$M=\frac{17.6g*\frac{1mol}{36.46g} }{14.5L} \\\\M=0.0333 M$

Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:

$pH=-log(0.0333)\\\\pH=1.477$

Which conserves as much decimals as significant figures in the molarity.

Learn more about pH calculations: brainly.com/question/1195974

3 0

2 years ago

The reaction 2PH3(g)+As2(g)⇌2AsH3(g)+P2(g) has Kp=2.9×10−5 at 873 K. At the same temperature, what is Kp for each of the followi

slavikrds [6]

Answer:

Part A

Kp = 3.4 x 10⁴

Part B

Kp = 2.4 x 10⁻¹⁴

Part C

Kp = 1.2 x 10⁹

Explanation:

2PH₃(g) + As₂(g) ⇌ 2 AsH₃(g) + P₂(g) Kp = 2.9 x 10⁻⁵

Kp = [AsH₃]²[P₂]/[PH₃]²[As] = 2.9 x 10⁻⁵

Part A

it is the inverse of the equilibrium given

Kp(A) = 1/ Kp = 1 / 2.9 x 10⁻⁵ = 3.4 x 10⁴

Part B

Is the equilibrium where the coefficients have been multiplied by 3,

Kp(B) = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴

Part C

This is the reverse equilibrium multipled by 2.

Kp(C) = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹

8 0

3 years ago

4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T

andre [41]

Answer: The enthalpy change of the reaction is -243 J/mol

Explanation:

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

$q=c\times \Delta T$

where,

c = heat capacity of calorimeter = 1620 J/K

$\Delta T$ = change in temperature = $0.150^oC=0.150K$ (Change remains same)

Putting values in above equation, we get:

$q=1620J/K\times 0.15K=243J$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}$

We are given:

Moles of monoclinic sulfur = 1 mole

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -243 J

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol$

Hence, the enthalpy change of the reaction is -243 J/mol

8 0

3 years ago

How many bases are needed to code for one amino acid?

N76 [4]

Answer:

Three. Also known as a triplet code.

4 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

What should be the labels on the X-axis and Y-axis?​

What should be the labels on the X-axis and Y-axis?