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Andru [333]
3 years ago
12

If a 0.4681 g Mg strip reacts with 0.650 M HCl in a 139.3 mL flask at 25oC, what is the minimum volume (mL) of HCl needed to com

pletely react with the magnesium?
Chemistry
1 answer:
Marizza181 [45]3 years ago
7 0
The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,

    Mg + 2HCl --> H2 + MgCl2

The number of moles of HCl that is needed for the reaction is calculated below.
    n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
    n = 0.0385 mols HCl

From the given concentration, we calculate for the required volume. 
    V = 0.0385 mols HCl/(0.650 mols/L)
     V = 0.05926 L or 59.26 mL

<em>Answer: 59.26 mL of HCl</em>
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one side of a cube measures 0.53 cm. the mass of the cube is 0.92 g. what is the density of the cube
ziro4ka [17]

Answer:

6.17 g/cm³

Explanation:

Data given:

one side of cube = 0.53 cm

mass of the cube is 0.92 g

density of the cube = ?

Solution:

First we will calculate for volume the cube

As we know all the sides or edges of a cube are equal so volume equation will be

So,

    V = length x width x height

    V = e³

as on side = 0.53 cm

then

     V = (0.53 cm)³

     V = 0.149 cm³

Now we will calculate density of cube

To calculate density, formula will be used

             d = m/v . . . . . (1)

where

d = density

m = mass

v = volume

put values in above formula 1

                   d = 0.92 g / 0.149 cm³

                   d = 6.17 g/cm³

so. the density of cube = 6.17 g/cm³

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3 years ago
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Explanation:

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Answer:

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3 years ago
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