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3 years ago

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Which atom of water is associated with the cation? Which atom of water is associated with the cation? Oxygen atom, because the p

artial negative charge is localized on the oxygen atom. Hydrogen atom. Since the number of hydrogen atoms is twice the number of oxygen atoms, it is more probable for the cation to be associated with the hydrogen atom.

1 answer:

Talja [164]3 years ago

4 0

Answer:

Oxygen atom, because the partial negative charge is localized on the oxygen atom.

Explanation:

A cation is a positive ion, so to associate with it, would be necessary a negative ion, an anion. When the water dissociates, it will form $2H^{+}$ and $O^{-2}$ .

So, oxygen has a negative charge, and in the molecule, the partial negative charge because it's more electronegative. Then, oxygen will be associated with the cation.

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Use the following balanced equation to answer the questions below.

Blababa [14]

Answer:

A. 4.5 mol Mg(OH)₂

B. 6 mol NaOH

Explanation:

Let's consider the following balanced equation.

Mg(NO₃)₂ + 2 NaOH ⇒ Mg(OH)₂ + 2 NaNO₃

PART A

The molar ratio of NaOH to Mg(OH)₂ is 2:1. The moles of Mg(OH)₂ produced from 9 moles of NaOH are:

9 mol NaOH × 1 mol Mg(OH)₂/2 mol NaOH = 4.5 mol Mg(OH)₂

PART B

The molar ratio of NaOH to NaNO₃ is 2:2. The moles of NaOH needed to produce 6 moles of NaNO₃ are:

6 mol NaNO₃ × 2 mol NaOH/2 mol NaNO₃ = 6 mol NaOH

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3 years ago

Which statements accurately describe elements? Check all that apply.

svet-max [94.6K]

Answer:

Elements are made up of only one type of atom.
Each element has a unique chemical symbol.
Elements can be identified by their atomic number.

Explanation:

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2 years ago

Read 2 more answers

Why is air considered a mixture not a compound?

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<span>The best reason I can think of for why we believe that air is a mixture is that freezing air slowly yields different liquids at different temperatures. Liquid nitrogen has a different boiling point than liquid oxygen. They also freeze at different temperatures. If air were only 1 compound, then air in its entirety would have a single boiling point and a single freezing point. </span>

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3 years ago

Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con

arsen [322]

Answer:

a) $T_b=590.775k$

b) $W_t=1.08*10^4J$

d) $Q=3.778*10^4J$

d) $\triangle V=4.058*10^4J$

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

$t = 20+273$

$t = 293k$

Initial heat $Q=1.36 * 10^4 J$

a)

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

$T_b-293k=297.775$

$T_b=590.775k$

b)

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

$PV=Wbc$

$Wbc=(2.20)(8.314)(1181_590.778)$

$Wbc=10805.7J$

Total Work-done $W_t$

$W_t=Wab+Wbc$

$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

c)

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

d)

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

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