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Ivanshal [37]
3 years ago
9

Based on Reference Table I, which change occurs when pellets of solid NaOH are added to water and stirred?

Chemistry
2 answers:
Alex Ar [27]3 years ago
8 0
The correct answer is option 1. When pellets of solid NaOH are added to water and stirred, the temperature of water increases because chemical energy is converted into heat energy. Dissolution of sodium hydroxide in water is an exothermic process where heat is dissipated.
kicyunya [14]3 years ago
3 0

The correct answer is option 1, that is, the temperature of the water increases as chemical energy gets transformed into heat energy.  

When the pellets of sodium hydroxide are mixed and stirred in water, the dilution of sodium hydroxide takes place, and then it gets dissociated into ions, the whole process is exothermic. The dissolution of sodium hydroxide is tremendously exothermic, the chemical burns as well as thermal burns can take place.  

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A rectangle has a volume of 395cm3. It has a mass of 147g. What is its density? *
LekaFEV [45]

The density of a rectangle : ρ = 0.372 g/cm³

<h3>Further explanation</h3>

Given

The volume of rectangle : 395 cm³

Mass : 147 g

Required

The density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume

Input the value :

ρ = 147 g : 395 cm³

ρ = 0.372 g/cm³

6 0
3 years ago
. How many milliliters of 0.20 M HCl are needed to exactly neutralize 40. milliliters of 0.40 M KOH
Anuta_ua [19.1K]

Answer:

V_{HCl}=80mL

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

KOH+HCl\rightarrow KCl+H_2O

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

n_{HCl}=n_{KOH}\\\\M_{HCl}V_{HCl}=M_{KOH}V_{KOH}

Hence, we compute the volume of HCl as shown below:

V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL

Best regards.

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