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2 years ago

An ate or ite at the end of a compound name usually indicates that the compound contains___________.

Chemistry

2 answers:

elena-14-01-66 [18.8K]2 years ago

7 0

Answer:

The correct answer is option d. a polyatomic anion.

Explanation:

Hello! Let's solve this!

The ending "ate" or "ite" indicates that there is a negative polyatomic ion (polyatomic anion). This means that the termination of the name will indicate the valence number of the element, if the number used is the highest or the lowest.

An example is:

CO-3: is the carbonate ion

SO-4: is the sulfate ion

We conclude that the correct answer is option d. a polyatomic anion.

Ymorist [56]2 years ago

3 0

Answer:

Explanation:

According to IUPAC nomenclature, a negative polyatomic ion has its name ending in 'ate' or 'ite'

As in carbonate (CO3^2-)

Arsenite (AsO3-)

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HELP ME PLEASEEE- 8th grade science

icang [17]

Answer:

It's obviously LIGHT ENERGY for me cuz it's from the flashlight..

Explanation:

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3 0

1 year ago

What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.1

Yuri [45]

Answer:

The minimum pressure should be 901.79 kPa

Explanation:

Step 1: Data given

Temperature = 25°C

Molarity of sodium chloride = 0.163 M

Molarity of magnesium sulfate = 0.019 M

Step 2: Calculate osmotic pressure

The formula for the osmotic pressure =

Π=MRT.

⇒ with M = the total molarity of all of the particles in the solution.

⇒ R = gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25 °C = 298 K

NaCl→ Na+ + Cl-

MgSO4 → Mg^2+ + SO4^2-

M = 2(0.163) + 2(0.019 M)

M = 0.364 M

Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)

Π = 8.90 atm

(8.90 atm)(101.325 kPa/atm) = 901.79 kPa

The minimum pressure should be 901.79 kPa

6 0

2 years ago

Part 1: A cylinder containing 20.0 L of compressed nitrogen is connected to an empty (evacuated) vessel with an unknown volume.

baherus [9]

Answer:

The volume of the vessel is 250 L

Partial pressure of hydrogen = 189 torr

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 20.0 L

V₂ = ?

P₁ = 25 atm

P₂ = 2 atm

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${25}\times {20.0}={2}\times {V_2}$

${V_2}=\frac {{25}\times {20.0}}{2}\ L$

${V_2}=250\ L$

The volume of the vessel is 250 L.

According to Dalton's law of partial pressure:-

$P_{H_2}=Mole\ fraction\times Total\ Pressure$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ H_2=\frac {n_{H_2}}{n_{H_2}+n_{He}}$

Also,

Mole fraction of H₂ = 1 - Mole fraction of He = 1 - 0.75 = 0.25

So,

Total pressure = 756 torr

Thus,

$P_{H_2}=0.25\times 756\ torr$

Partial pressure of hydrogen = 189 torr.

7 0

2 years ago

For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.

Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0

2 years ago

Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water i

gtnhenbr [62]

Answer:

The percentage of the void space out of the total volume occupied is 93.11%

Explanation:

A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.

To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom

Let's calculate this one after the other.

For the hydrogen, formula for the volume will be

$V_{hydrogen$ = 2 × 4/3 × π × $r_{H}^{3}$

where $r_{H}^{3}$ = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × $10^{-12}$ meters

Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31

we multiply this by the avogadro's number = 6.02 * 10^23

= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3

We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen

Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3

adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)

Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water

= (18g/mol)/(1 g/ml) = 18 ml/mol

Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml

Now, the percentage of void = volume of void/total volume * 100%

= 16.76/18 * 100% = 93.11%

7 0

2 years ago