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2 years ago

10

Which types of bond is formed between the two chlorine atoms in a chlorine molecule?

1 answer:

madam [21]2 years ago

7 0

They achieve stable structures by sharing their single, unpaired electron.

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Name the compound that makes up the white ring

Brut [27]

Answer:

ammonium chloride

Explanation:

if it's the white ring I think your talking abt

3 0

2 years ago

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Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): 2H2O2

zalisa [80]

Answer:

23.0733 L

Explanation:

The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

$Mass=\frac {50}{100}\times 125\ g$

Mass = 62.5 g

Molar mass of $H_2O_2$ = 34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{62.5\ g}{34\ g/mol}$

$moles= 1.8382\ mol$

Consider the given reaction as:

$2H_2O_2_{(aq)}\rightarrow2H_2O_{(l)}+O_2_{(g)}$

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.

Also,

1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.

So,

1.8382 moles of hydrogen peroxide decomposes to give $\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr
The conversion of P(torr) to P(atm) is shown below:
[tex]P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 746 / 760 atm = 0.9816 atm

Temperature = 27 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (27 + 273.15) K = 300.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K

<u>⇒V = 23.0733 L</u>

8 0

2 years ago

Pls help! I'll give u 10 points and the brainlest answer!!!

alekssr [168]

Answer:

371km

Explanation:

the lower the more fuel there is

7 0

3 years ago

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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

If a reaction taking place in a flask shows a drop in temperature, which type of reaction is it? Options: exothermic reactions e

liubo4ka [24]

Answer:

exothermic reaction

Explanation:

If there is a drop in temperature, then energy was lost to the surroundings because temperature is the average measure of kinetic energy. An exothermic reaction would result in this lost of energy. An endothermic reaction would absorb energy and make the temperature rise.

4 0

3 years ago

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