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1 year ago

This is like 40 points!! Pls help.

Chemistry

1 answer:

abruzzese [7]1 year ago

4 0

Answer:

When meteoroids enter Earth's atmosphere (or that of another planet, like Mars) at high speed and burn up, the fireballs or “shooting stars” are called meteors. When a meteoroid survives a trip through the atmosphere and hits the ground, it's called a meteorite.

Explanation:

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Which step will decrease the pressure of a gas inside a closed cubical container? increasing the number of moles of gas decreasi

Jobisdone [24]

Answer: Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.

Explanation:

According to Gay-Lussac's Law : 'The pressure of the gas increases with increase in temperature of the gas when volume of the gas is kept constant'.

$(Pressure)\propto (Temperature)$

At constant volume, pressure of the gas will decrease on decreasing the temperature or vice versa.

Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.

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2 years ago

Read 2 more answers

The azide ion, n−3, is a symmetrical ion, all of whose contributing structures have formal charges. draw three important contrib

stich3 [128]

Explanation:

Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.

The azide ion that is $N^{-}_3$ , is a symmetrical ion, all of whose contributing structures have formal charges.

Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.

Contributing structures of azide ion are drawn in the image attached.

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2 years ago

An aqueous solution has a volume of 2.0 L and contains 36.0g of glucose. If the Molar Mass of glucose is 180g/mol, what is the m

choli [55]

Answer:

moles = 36/180 = 0.2 moles

molarity = 0.2/2 = 0.1 mol/dm3

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2 years ago

Which type of internal structure represents a gemstone.

Lemur [1.5K]

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2 years ago

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

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2 years ago