2 years ago

Hey can someone help me figure out what ph of ajax cleaner and what color it changes to?

Chemistry

1 answer:

serious [3.7K]2 years ago

8 0

I just know the ph is between 7 and 8

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A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

S = $K_{b} \times ln \Omega$

where, S = entropy

$K_{b}$ = Boltzmann constant

$\Omega$ = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that $\Omega$ = 1, and $\Omega'$ = 0.833

Therefore, change in entropy will be calculated as follows.

$\Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega$

= $1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)$

= $1.38 \times 10^{-23} \times (-0.182)$

= $-0.251 \times 10^{-23}$

or, = $-2.51 \times 10^{-24}$

Thus, we can conclude that the entropy change for a particle in the given system is $-2.51 \times 10^{-24}$ J/K particle.

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2 years ago

Calculate no. Of moles in 329g of H2so4

Anestetic [448]

Mol= mass (grams) /Mr
Mr of Sulfuric Acid (H2SO4): 98
mol= 329/98
=3.36 moles

8 0

2 years ago

The oxidation number of Na in NaCl a. 0 b. -1 c. +1 d. -2 e. +2

Dahasolnce [82]

NaCl:

Na = + 1

Cl = - 1

hope this helps!

4 0

3 years ago

The half reaction of an oxidation-reduction shows that iron gains electrons. What does this electron gain mean or iron?

ale4655 [162]

Answer:

C. its reduced

Explanation:

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2 years ago

Uranium-236 is _____. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-

nata0808 [166]

<span>Uranium-236 is intermediate nuclei. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-bomb.

Following is the reaction involved in above process:

</span> $^{235}U$ + $^{1}n$ → $^{236}U$ → $^{144}Ba$ + $^{89}Kr$ + 3 $^{1}n$ <span> + 177 MeV
</span>
Here,
$^{235}U$ = Fission material,
$^{1}n$ = projectile,
$^{236}U$ = intermediate nuclei,
$^{144}Ba$ and $^{89}Kr$ = Fission product

8 0

2 years ago

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