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swat32
3 years ago
9

Se mezclan por error 70 mililitros de solución de HCl 0.5 Normal con 250 mililitros de solución de hidróxido de sodio 0.30 norma

l. Calcular: El volumen final de la solución resultante El carácter de la solución resultante La concentración final de la solución resultante El pH de la solución resultante El valor del pOH de la solución resultante
Chemistry
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

asxrsse367721479009qqucinübjpnpnfzdh ilo h

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No, tobacco companies do not reuse tobacco in their cigarettes, I also did some extra research to make sure I wasn’t giving a false answer

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3 years ago
Which substance has the highest viscosity?
Lemur [1.5K]
Room temperature has the highest viscosity
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What is the correct name for the compound named 1-chloro-3-pentyne
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5-chloro-2-pentyne. The chain must be numbered starting with the end nearest the triple bond.
7 0
3 years ago
For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)
juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = <u>4.82 grams benzene</u>

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Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

<u>Mass CO2 = 9.15 grams</u>

4 0
3 years ago
Look at the following equation.
Ronch [10]

Answer:

A

Explanation:

_N2 + _H2 - _NH3

LHS RHS

N=2×1=2 N=1×2=2

H=2×3=6 H=3×2=6

1,3,2

4 0
3 years ago
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