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3 years ago

In the reaction 4Al+_O2→2Al2O3 , what coefficient should be placed in front of the o2 to balance the reaction?

Chemistry

2 answers:

deff fn [24]3 years ago

7 0

Answer:

yeep the answer is 3... i just took the k12 test rn

DIA [1.3K]3 years ago

4 0

Answer:

Explanation:

took the test

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Rank the following bonds in order of decreasing bond length, based on periodic trends, starting with the longest bond at the top

sattari [20]

Answer:

H-BI,H-Se,H-S,H-I,H-Br

Explanation:

One thing that must be kept in mind is that atomic size increases down the group and decreases across the period. The bond lengths of species are influenced by the relative sizes of atoms or ions present in the bond.

The bonds in the answer have been arranged on basis of their decreasing atomic size because the greater the atomic size of the atoms, the greater the bond length and vice versa.

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4 years ago

In a double-blind test, the non-control group receives a

Aleksandr [31]

Answer:

Hi There! I'm new so I hope I get This Right! ^^

Explanation:

double blind test. the control group receives a placebo. Why is a placebo used in a double-blind test? so that the effects on people in two different groups can be compared.

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4 years ago

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

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3 years ago

What new scientific tool is very useful to observe changes in Landforms?

evablogger [386]

I’m not sure but I think it’s Earthquake Detector

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3 years ago

Read 2 more answers

How many moles each element are in a mole of methane?

german

$\large \mathcal{ANSWER}$

This means that the mass of one methane molecule is 12.011 u + (4 × 1.008u), or 16.043 u. This means that one mole of methane has a mass of 16.043 grams. A mole can be thought of as two bags of different sized balls. One bag contains 3 tennis balls and the other 3 footballs.

Samples of helium (He), nitrogen (N 2 ), and methane (CH 4 ) are at STP. Each contains 1 mole or 6.02 × 10 23 particles. However, the mass of each gas is different and corresponds to the molar mass of that gas: 4.00 g/mol for He, 28.0 g/mol for N 2 , and 16.0 g/mol for CH 4 .

3 0

3 years ago

In the reaction 4Al+_O2→2Al2O3 , what coefficient should be placed in front of the o2 to balance the reaction?​

In the reaction 4Al+_O2→2Al2O3 , what coefficient should be placed in front of the o2 to balance the reaction?