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2 years ago

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Which is more likely HON or HNO

1 answer:

netineya [11]2 years ago

6 0

NO- (isoelectronic or same as O2) is hard base

H+ is hard acid

bonding depends which pi* has more electron density and calc shows more on N...hence HNO is more likely (bent) molecule

whatever....

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K2so4 is a strong electrolyte. determine the concentration of each of the individual ions in a 0.450 m k2so4 solution.

amid [387]

<em>Answer:</em>

Conc. of K+ ions = 0.90 M
Coc. of SO4∧-2 = 0.45 M

<em>Explanation:</em>

<em>Data Given:</em>

Conc. of H2SO4 = 0.450

As sulphoric acid is a strong electrolyte, it completely dissociate into ions.

H2SO4 ⇆ 2K+ + SO4∧-2

.450 M K2SO4 means that there is .450 mols of K2SO4 in every liter of solution.

K2SO4 : K+ K2SO4 : SO4∧-2

1 = 2 1 = 1

0.450 = 2× 0.450 = 0.90 0.450 = 0.450×1 = 0.450

<em> Result:</em>

Conc. of potassium ion will be 0.90M

Coc. of sulphate ions will be 0.45 M

8 0

3 years ago

Bronze is an alloy made of mostly copper (Cu) and tin (Sn). What type of bond forms bronze?

Snezhnost [94]

Answer:

A. Metallic bond

Explanation:

Think about it: copper and tin are both common metals. That's how we know it's a metallic bond!

Why not B: Covalent bonds are between two nonmetals.

Why not C: Ionic bonds are between a nonmetal and a metal.

Why not D: Paired bond isn't a common phrase in chemistry.

6 0

3 years ago

If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?

Yuki888 [10]

Answer:

›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element

Explanation:

5 0

2 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

HELP ME PLS....... PLSSSSSSSSSSSSSSSSSSSSSS

Tems11 [23]

Answer: A different group of scientists using different methods.

5 0

2 years ago