Please help me I don’t know this

Which methods could be used to dilute a solution of sodium chloride (NaCl)? Select all that apply.

Pavel [41]

Answer:

NaOH₂2CO3

Explanation:yes i know it is not a Balinese equation

4 0

3 years ago

(100 POINTS) Write the chemical formula of the ionic compound magnesium chlorate. The magnesium (Mg) ion has a +2 charge. Chlora

antoniya [11.8K]

Answer:

A

Explanation:

since Mg has a charge of +2 and ClO3 has a charge of 1-, you need 2 ClO3 to cancel out the +2 since 2 ClO3 ions would have a 2- charge

6 0

3 years ago

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

7 0

3 years ago

When parallel light rays exit a convex lens, they .

Alexxx [7]

When the parallel light rays exit a convex lens, it comes away or diverge.

6 0

3 years ago

Read 2 more answers

Calculate the Lattice Energy of KCl(s) given the following data using the Born-Haber cycle: ΔHsublimation K = 79.2 kJ/mol IE1 K

DanielleElmas [232]

Answer:

Explanation:

The following equation relates to Born-Haber cycle

$\triangle H_f = S + \frac{1}{2} B + IE_M - EA_X+ U_L$

Where

$\triangle H_f$ is enthalpy of formation

S is enthalpy of sublimation

B is bond enthalpy

$IE_M$ is ionisation enthalpy of metal

$EA_X$ is electron affinity of non metal atom

$U_L$ is lattice energy

Substituting the given values we have

-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L

$U_L$ = - 707 KJ / mol

3 0

3 years ago