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MAVERICK [17]
2 years ago
14

What is the concentration of hcl in the final solution when 65 ml of a 12 m hcl solution is diluted with pure water to a total v

olume of 0.15 l?
Chemistry
1 answer:
Gre4nikov [31]2 years ago
6 0
<span>First I converted 0.15L in to mL (150mL) Then, (1.5M)*(65mL)=X *(150mL) (1.5M)*(64mL)/(150mL)=0.65M Answer;3.4 x 10^-2 M</span>
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sweet [91]

Answer:

no questions

it's just all dots here

5 0
2 years ago
Which of the following is the term used to describe a set of elements whose properties tend to be largely predictable based on p
lys-0071 [83]
Group.... because elements in a group have same chemical properties
8 0
2 years ago
Read 2 more answers
If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degrees C, what is the mixture of gasses?
jonny [76]

Answer:

The total pressure of the gas mixture = 2.76 atm

Note: The question is not complete. The complete question is as follow:

If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degree Celsius , what is the pressure of the mixture of gases?

Explanation:

Mass of O₂ gas = 43.1 g, molar mass of O₂ gas = 32.0 g/mol

Number of moles of O₂ gas = 43.1/32.0 = 1.347 moles

Mass of CO₂ gas = 6.8 g, molar mass of CO₂ gas = 44.0 g

Number of moles of CO₂ gas = 6.8/44 = 0.155 moles

Total number of moles of gas mixture, n = (1.347 + 0.155) = 1.502 moles

Volume of gas mixture, V = 13.7 L

Temperature of gas mixture, T = 34 °C = (273.15 + 34) K = 307.15 K

Pressure of gas mixture = ?

Molar gas constant, R = 0.0821 liter·atm/mol·K.

Using the ideal gas equation: PV =nRT

P = nRT/V

P = (1.502 × 0.0821 × 307.15) / 13.7

P = 2.76 atm

Therefore, the total pressure of the gas mixture = 2.76 atm

5 0
2 years ago
The total volume in milliliters of a glucose-water solution is given by the equation below: V = 1001.93 + 111.5282m + 0.64698m^2
vagabundo [1.1K]

Answer:

a. Vₐ = 111.5282 + 1.29396m

b. For m = 0.100m; Vₐ = 111.6576

Explanation:

The partial molar volume of compound A in a mixture of A and B is defined as :

V_a = \frac{dV}{dn_a}

Where V is volume and n are moles of a.

a. As molality is proportional to moles of substance, partial molar volume of glucose can be defined as:

Vₐ =  dV / dm =  d(1001.93 + 111.5282m + 0.64698m²) / dm

<em>Vₐ = 111.5282 + 1.29396m</em>

b. Replacing for m = 0.100m:

Vₐ = 111.5282 + 1.29396×0.100

<em>Vₐ = 111.6576</em>

<em></em>

I hope it helps!

3 0
3 years ago
For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in th
CaHeK987 [17]

Answer:

a. 7278 K

b. 4.542 × 10⁻³¹

Explanation:

a.

Let´s consider the following reaction.

N₂(g) + O₂(g) ⇄ 2 NO(g)

The reaction is spontaneous when:

ΔG° < 0  [1]

Let's consider a second relation:

ΔG° = ΔH° - T × ΔS° [2]

Combining [1] and [2],

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)

T >  7278 K

b.

First, we will calculate ΔG° at 25°C + 273.15 = 298 K

ΔG° = ΔH° - T × ΔS°

ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K

ΔG° = 173.1 kJ/mol

We can calculate the equilibrium constant using the following expression.

ΔG° = - R × T × lnK

lnK = - ΔG° / R × T

lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K

K = 4.542 × 10⁻³¹

7 0
2 years ago
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