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USPshnik [31]
3 years ago
14

Compounds with very high vapor pressures must have very minimal intermolecular forces true or false

Chemistry
1 answer:
g100num [7]3 years ago
5 0
It's true. Compounds have very weak ones the higher the vapor pressure. Hope that answerd your question ☺️☺️
You might be interested in
Codons.
andrey2020 [161]

Answer:

1. C- Three.

2. A- Methionine

3. D- Translocation.

4. C- OH.

5. A - 5'

6. A - 3' carbon

7. A. adenine and guanine

Explanation:

1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.

2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.

3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.

4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).

5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.

6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.

7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).

7 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
What is the best description of weathering?
zheka24 [161]
D. breakdown of rocks through mechanicals or chemicals processes
8 0
3 years ago
How will you ensure your server is protected from data hacks?<br>​
mario62 [17]
Close all network ports, filter those you can't block

Firewalls are used to filter network traffic and are available as standard system software on most operating systems. Limit the openings hackers have to your server.
4 0
2 years ago
Read 2 more answers
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid. HCl(aq), as described by
melisa1 [442]

Answer:

0.88 g

Explanation:

Using ideal gas equation to calculate the moles of chlorine gas produced as:-

PV=nRT

where,

P = pressure of the gas = 805 Torr

V = Volume of the gas = 235 mL = 0.235 L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 62.3637\text{torr}mol^{-1}K^{-1}

n = number of moles of chlorine gas = ?

Putting values in above equation, we get:

805torr\times 0.235L=n\times 62.3637\text{ torrHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{805\times 0.235}{62.3637\times 298}=0.01017\ mol

According to the reaction:-

MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2

1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.

So,

0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.

Moles of MnO_2 = 0.01017 moles

Molar mass of MnO_2 = 86.93685 g/mol

So,

Mass=Moles\times Molar\ mass

Applying values, we get that:-

Mass=0.01017moles \times 86.93685\ g/mol=0.88\ g

<u>0.88 g of MnO_2(s) should be added to excess HCl (aq) to obtain 235 mL of Cl_2(g) at 25 degrees C and 805 Torr.</u>

6 0
3 years ago
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