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lara31 [8.8K]
3 years ago
10

Distilled water will

Chemistry
1 answer:
Kruka [31]3 years ago
6 0
The answer is (c) as distilled water has the same pH as normal water therefore it is not able to react with any litmus paper.
Hope this helps :).
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What do the spectral lines represent?
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Question

What do the spectral lines represent?

Answer:

Spectral lines is defined by quantum mechanics in terms of energy levels of ions, molecules, and atoms. When electrons move from a higher level to lower one, photons are emitted and an emission line can be seen in the spectrum. As electrons jump down to the n=2 orbit, they emit photons of specific frequency (hence colour) that can be seen as emission lines in the visible part of spectrum.

Explanation:

8 0
3 years ago
Calculate the pH of a buffer that is 0.020 M HF and 0.040 M NaF. The Ka for HF is 3.5 × 10-4.
zzz [600]
 <span>pH = pKa + log ([base]/[acid]) = -log (3.5 x 10^-4) + log (0.040/.020) = 3.46 - 0.30 = 3.16</span>
4 0
4 years ago
Read 2 more answers
Will favor dissolution and
Marizza181 [45]

Answer:

b) heating, cooling

Explanation:

Heating will favour dissolution (it helps dissolving faster).

Cooling helps in crystallization.

8 0
2 years ago
What happens to the air temperature of a decending mass of air
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Hope this helps
4 0
3 years ago
Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)
Sedbober [7]

Answer:

The yield of the product in gram is \mathsf{{w_P}=0.26 \ gram}

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = 2.7 \times 10^{-3} \ moles

We know that:

Number of moles = mass/molar mass

Then:

2.7 \times 10^{-3} = \dfrac{ mass}{257.997}

mass = 2.7 \times 10^{-3} \times 257.997

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = \dfrac{n_P}{n_R}\times 100\%

where;

n_P = \dfrac{w_P}{m_P}    and n_R = \dfrac{w_R}{m_R}

Theoretical yield = \dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

Given that the theoretical yield = 100%

Then:

100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}

{w_P}=\dfrac{w_R \times m_P}{m_R}

where,

w_P = derived weight of the product

m_P =the molecular mass of the derived product

m_R = the molecular mass of the reagent

w_R = weight in a gram of the reagent

{w_P}=\dfrac{w_R \times m_P}{m_R}

{w_P}=\dfrac{0.697 \times 96.2}{257.997}

\mathsf{{w_P}=0.26 \ gram}

8 0
3 years ago
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