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kramer
2 years ago
13

Given one mole of each substance, which of the following will produce the FEWEST particles in aqueous solution? 1. sodium nitrat

e 2. CH2Cl2 3. sodium phosphate 4. K2SO4
Chemistry
1 answer:
KonstantinChe [14]2 years ago
7 0

Answer:

CH2Cl2 produces the fewest particles

Explanation:

Step 1: Data given

1. sodium nitrate = NaNO3

2. CH2Cl2

3. sodium phosphate = Na3PO4

4. K2SO4

Step 2

NaNO3 → Na+ +NO3-

1. Sodium nitrate (NaNO3) dissociates into one Na^+ ion and one nitrate (NO3^-) ion. In total we have <u>2 particles</u>.

2. CH2Cl2 is a covalent compound, so, it doesn't dissociate. So, <u>one particle</u>.

3) Na3PO4 → 3Na+ + PO4^3-

Sodium phosphate (Na3PO4) is  an ionic compound. It dissociates into 3 Na^+ ions and 1 phosphate(PO4^-3) ion. In total, <u>4 particles. </u>

4) K2SO4 → 2K+ + SO4^2-

Because K2SO4 is an "ionic" compound, it dissociates into 2 K^+ ions, and one sulfate(SO4^-2) ion. So, <u>3 particles</u>.

CH2Cl2 produces only 1 particle so it has the fewest particles

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A rod, X has a positive charge of 8. An otherwise identical rod, Y has a negative charge of 4. The rods are touched together, an
gtnhenbr [62]

Answer:

1.  electrons

2. From "Y" to "X"

Explanation:

1. Electrons move between the rod since the electrons are the only charge carriers which are free to move.

2. The particles move from from "Y" to "X"  since the electrons are the only charge carriers which are free to move. The positive charge on rod x is due to a deficit of electrons while the negative charge on rod Y is due to the excess of electrons. When the rods come together, the electrons move from "Y" to "X" since the electrons are the only charge carriers which are free to move.

4 0
1 year ago
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
choose the reaction that illustrates delta H *f for Ca(NO3)2.(A) Ca (s) + N2 (g) + 3O2 ---&gt; Ca(NO3)2 (s)(B) Ca2 (aq) + 2 NO3-
kompoz [17]

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Standard enthalpy of formation is the change in enthalpy of  one mole of a substance present at the standard state that is 1 atm of pressure and 298 K of temperature. The substance is formed from its pure elements under the same conditions.

We are given a chemical compound having chemical formula Ca(NO_3)_2

This compound is formed by the combination of calcium, nitrogen and oxygen elements.

The chemical equation for the formation of Ca(NO_3)_2 from the components in their standard states follows:

Ca+N_2+3O_2\rightarrow Ca(NO_3)_3

Hence, the correct answer is Option A.

3 0
2 years ago
trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is
spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 cm^{-1}.

Explanation:

Bands that appear above 3000 cm^{-1}  are often unsaturation diagnoses suggest. The band at 3000- 3100 cm^{-1} is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

4 0
2 years ago
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goldenfox [79]

Answer:

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Explanation:

when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

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3 0
2 years ago
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