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PIT_PIT [208]
2 years ago
15

State the color of methyl orange in a sample of NaOH

Chemistry
1 answer:
devlian [24]2 years ago
5 0
I just need points I hate my sister
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If Log 4 (x) = 12, then log 2 (x / 4) is equal to
Alexus [3.1K]

The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

              = log₂(2²⁴/2²)

              = log₂(2²⁴ ⁻ ²)

              = log₂(2²²)

On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

We know that logₐa = 1;

So,

log₂(x/4) = 22(1)

∴ log₂(x/4) = 22.

Learn more about the properties of logarithm here:

brainly.com/question/12049968

#SPJ9

8 0
1 year ago
A gas at 300 k and 4.0 atm is moved to a new loacation with a temperature of 250 k. The volume changes from 5.5 L to 2.0 L. What
alexgriva [62]
Answer is: <span>the pressure of the gas is 9,2 atm.
</span>p₁ = 4,0 atm.
T₁ = 300 K.
V₁ = 5,5 L.
p₂ = ?
T₂ = 250 K.
V₂ = 2,0 L.
Use combined gas law - the volume of amount of gas is proportional to the ratio of its Kelvin temperature and its pressure.<span> 
</span>p₁V₁/T₁ = p₂V₂/T₂.
4 atm · 5,5 L ÷ 300 K  = p₂ · 2,0 L ÷ 250 K.
0,0733 = 0,008p₂.
p₂ = 9,2 atm.

4 0
3 years ago
How can one describe the bond between the two atoms of hydrogen to the one atom of oxygen within a molecule of water (h2o)?
Digiron [165]

There are O-H bonds in H2O. They have the intramolecular force of polar covalent bond.

4 0
3 years ago
Sulfuric acid is a component of acid rain formed when gaseous sulfur dioxide pollutant reacts with gaseous oxygen and liquid wat
Vesna [10]
2SO₂(g) + O₂(g) + 2H₂O(aq) → 2H₂SO₄(aq)
6 0
3 years ago
Read 2 more answers
Chemistry Lab Question: Filtration
tatyana61 [14]

For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.

For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.

6 0
3 years ago
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