The smallest volume from among the following is

What are three main classifications of the elements

neonofarm [45]

Answer:

Metals, non metals and metalloid

Explanation:

4 0

4 years ago

Single stranded___ molecules translate he code for making protein

Radda [10]

Answer:

RNA

Explanation:

Central dogma is the process of the formation of protein from the sequence of nucleotide on DNA. It consists of two further processes Transcription and translation.

During Transcription, the information from the DNA sequence is transcripted in a sequence of RNA, that is single stranded and contains Uracil instead of Thymine. During Translation, the information from the RNA sequence is translated into a sequence of amino-acid, and a protein.

Hope it helps!

6 0

3 years ago

What does the most accurate clock tell us

Murrr4er [49]

It tell us the most accurate time u can get out of a clock

6 0

4 years ago

Read 2 more answers

Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freez

Elden [556K]

Answer:

The solutions are listed from the lowest freezing point to highest freezing point.

Mg₃(PO₄)₂ - AlBr₃ - BeBr₂ - KBr

Explanation:

Colligative property of freezing point, is the freezing point depression.

Formula is:

ΔT = Kf . m . i

Where ΔT = T° freezing pure solvent - T° freezing solution

Kf is the cryoscopic constant

m is molality and i is the Van't Hoff factor (number of ions, dissolved in solution)

In this case, T° freezing pure solvent is 0° because it's water, so the Kf is 1.86 °C/m, and m is 0.1 molal.

The i modifies the T° freezing solution. The four salts, are ionic compounds, so for each case:

BeBr₂ → Be²⁺ + 2Br⁻ i = 3

AlBr₃ → Al³⁺ + 3Br⁻ i = 4

Mg₃(PO₄)₂ → 3Mg²⁺ + 2PO₄⁻³ i = 5

KBr → K⁺ + Br⁻ i = 2

Solution of Mg₃(PO₄)₂ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).

8 0

3 years ago

) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that t

sergeinik [125]

Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

$m_{calcite}=15.4448g-12.4631g=2.9817g$

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

$V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL$

Thus, the density of the calcite sample will be:

$\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL$

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

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Regards!

6 0

2 years ago