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kramer
3 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]3 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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Andrei [34K]

The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .

The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.

The thermochemical equation for the combustion of benzene is;

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol

We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.

Learn more: brainly.com/question/13164491

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2 years ago
What organelles are involved in the production of proteins
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endoplasmic reticulum (ER)

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3. What noble gas would be part of the electron configuration notation for Mn?
shusha [124]

Answer:

Argon {Ar}

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7 0
3 years ago
Use Hess's Law to calculate the enthalpy change for the reaction
Marysya12 [62]

Answer:

ΔH = 125.94kJ

Explanation:

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:

1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ

2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ

-1/2 (1):

WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ

3/2 (2):

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The sum of  last both reactions:

WO3(s) + 3H2(g) → W(s) + 3H2O(g)

ΔH = 842.7kJ -716.76kJ

<h3>ΔH = 125.94kJ </h3>
3 0
3 years ago
Find the mass of 175.4mL of benzene if the density is 0.8786g/ml
bulgar [2K]
Hey there!:

Volume = 175.4 mL

density = 0.8786 g/mL  

mass = ?

Therefore:

D = m / V

0.8786 = m / 175.4

m = 0.8786 * 175.4

m = 154.10644 g

hope this helps!


5 0
3 years ago
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