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3 years ago

Equivalent weight of reducing agent C2H2O4.2H2O + O= 2CO2 + 3H2O

Chemistry

1 answer:

astra-53 [7]3 years ago

7 0

Answer:

im too stoned for this

Explanation:

sorry man

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Chlorine forms a number of oxides with the following oxidation numbers: 1, 3, 4, 6, and 7. These compounds are stable and neutra

SashulF [63]

Answer:

$Cl_2O$ , $Cl_2O_3$ , $ClO_2$ , $ClO_3$ , $Cl_2O_7$

Explanation:

Empirical formula of the compound is the simplest ratio of elements present in the compound.

Empirical formula of compounds of chlorine with oxygen is as follows:

Compounds in which oxidation state of Cl is +1

$Cl_2O$

Compounds in which oxidation state of Cl is +3

$Cl_2O_3$

Compounds in which oxidation state of Cl is +4

$ClO_2$

Compounds in which oxidation state of Cl is +6

$ClO_3$

Compounds in which oxidation state of Cl is +7

$Cl_2O_7$

8 0

3 years ago

Please help

lora16 [44]

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3 years ago

What type of soil holds the least amount of water

jeka57 [31]

The clay soil holds the least amount if water.

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3 years ago

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During a laboratory experiment, 36.12 grams of Aboz was formed when O2 reacted with aluminum metal at 280.0 K and 1.4 atm. What

kvv77 [185]

Answer:

8.70 liters

Explanation:

3O₂+ 4Al → 2AI₂O₃

First we convert 36.12 g of AI₂O₃ into moles, using its molar mass:

36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃

Then we convert AI₂O₃ moles into O₂ moles, using the stoichiometric coefficients of the reaction:

0.354 mol AI₂O₃ * $\frac{3molO_2}{2molAl_2O_3}$ = 0.531 mol O₂

We can now use the PV=nRT equation to calculate the volume, V:

1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K

V = 8.708 L

5 0

3 years ago

A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il

Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

P= 2 atm
V= ?
n=0.223 moles
R= 0.0821 $\frac{L*atm}{mol*K}$
T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 $\frac{L*atm}{mol*K}$ * 306 K

Solving:

$V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\$

V= 2.80 L

The volume of the gas is 2.80 L.

7 0

3 years ago

Equivalent weight of reducing agent C2H2O4.2H2O + O= 2CO2 + 3H2O​

Equivalent weight of reducing agent C2H2O4.2H2O + O= 2CO2 + 3H2O