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3 years ago

"which compound would be expected to show intense ir absorption at both 3500 and 1733cm-1?"

1 answer:

4 0

The compound suppose to contain a carbonyl group which shows peak about 1735 Cm^-1 and it also contians an primary amino group for peak of 3500.

Carbonyl group is present in carboxyl group, ketones or ester.

Amino group in presence of carbonyl produces amide group

Thus, the compound must contain CONH_2 group

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What is the oxidizing agent in the reaction 2MnO4-(aq) 10I-(aq) 16H (aq) ---> 2Mn2 (aq) 5I2(aq) 8H2O(l)

faltersainse [42]

Correct Question: what is the oxidizing agent in the reaction.

2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Answer: MnO4-is the oxidizing agent

Explanation:

In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Oxidizing agent oxidizes other molecules while the themselves get reduced.

oxidizing agents give away Oxygen to other compounds.

MnO4-is the oxidizing agent because

On the reactants side

Oxidation number of Mn in 2MnO4- is +7

Oxidation number of Cl- is -1

On the products side

Oxidation number of Mn is +2

While oxidation number of Cl is zero

Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2

4 0

3 years ago

What is the correct balanced equation for the reaction of sodium with water?

Wittaler [7]

Answer:

A. 2Na + 2H₂O → 2NaOH + H₂

Explanation:

When sodium is reacted with water, a single replacement reaction occurs. The product of the reaction is typically sodium hydroxide and hydrogen gas.

Reactants:

Sodium + water

Product:

Sodium hydroxide + hydrogen gas

So;

2Na + 2H₂O → 2NaOH + H₂

6 0

2 years ago

A gas occupies 900.0 ML at temperatures of 27.0 Celsius. What is the volume at 123.0 celsius

Artemon [7]

Answer:

1188.0 mL.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 900 mL, T₁ = 27.0°C + 273 = 300.0 K.

V₂ = ??? mL, T₂ = 123.0°C + 273 = 396.0 K.

∴ V₂ = V₁T₂/T₁ = (900 mL)(396 K)/(300.0 K) = 1188.0 mL.

4 0

3 years ago

Read 2 more answers

A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by

Alla [95]

Answer: 1. The empirical formula is $PNF_2$

2. The molecular formula is $PNF_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P = $\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles$

Moles of F = $\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P = $\frac{1.20}{1.20}=1$

For N = $\frac{1.20}{1.20}=1$

For F = $\frac{2.41}{1.20}=2$

The ratio of P: N: F= 1: 1: 2

Hence the empirical formula is $PNF_2$

The empirical weight of $PNF_2$ = 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1$

The molecular formula will be= $1\times PNF_2=PNF_2$

3 0

2 years ago

If 88.0 L of natural gas, which is essentially methane (CH4), undergoes complete combustion at 720. mm Hg and 22ºC, how many gra

mr_godi [17]

126 grams of H2O is formed.

Explanation:

Data given:

volume of the gas = 88 Liters

pressure = 720 mm Hg or 0.947 atm

temperature T = 22 Degrees or 295.15 K

R = 0.08021 atm L/mole K

n =?

The formula is used is of ideal gas law to know the number of moles of CH4 undergoing combustion.

PV = nRT

n = $\frac{PV}{RT}$

putting the values in the equation

= 0.947 X 88/ 0.08021 X 295.15

n = 3.5 moles

balanced reaction for combustion of methane

CH4 + O2 ⇒ CO2 + 2H20

1 mole of CH4 undergoes combustion to form 2 moles of water

3.5 moles will give x moles of water

2/1 = x/3.5

x = 7 moles of water (atomic mass of water = 18 gram/mole)

mass = atomic mass x number of moles

mass = 18 x 7

=126 grams of water is formed.

7 0

3 years ago