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Juliette [100K]
3 years ago
7

Under identical conditions of temperature and pressure, which of the following gasses is the densest: Ne, CO2, or Cl2?

Chemistry
1 answer:
Tanzania [10]3 years ago
7 0
Cl2=3.17g/L
Ne=.901g/L
CO2=1.96g/l
 therefore Cl2 is the densest gas under the given conditions.
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An adult mosquitoes lifespan is about 16 days how many minutes is that
Delicious77 [7]
1 day = 1, 440 minutes
16 days = (1, 440 × 16) days = 23, 040 minutes
5 0
3 years ago
How many hydrogen atoms are attached to each carbon adjacent to a double bond? nurition?
larisa [96]
That depends. there are 2 possible answers.
      H
C - C = C - H gives a different answer on the right than on the left.

One the left side, the second Carbon is attached to a double bond and has but one hydrogen attached to it.

The Carbon on the right of the double bond has 2
     H
C- C = C - H
            H

I'm not sure what you should put. It's one of those things that I would repeat my argument and submit it.
3 0
4 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
3 years ago
Which of these pairs indicates an incorrect coupling of reversible reactions?dehydration synthesis and hydrolysisanabolic and ca
Olegator [25]

Answer:

option C= hydrolysis and break down

Explanation:

All other three pairs are correct coupling of each others.

Option A= dehydration synthesis and hydrolysis

Dehydration synthesis:

In dehydration synthesis monomers combine through the covalent bonds and form large molecules. The large molecules are called polymers. The water as a byproduct also released when monomers joints together.

Hydrolysis:

In hydrolysis the polymers are break down into monomers by using water molecules. The catalysts are also required in this process.

Option B= Catabolic and Anabolic

Anabolic:

In this process smaller molecules combine to gather to form large complex molecules by using energy.

For example simple glucose molecules join together to form large disaccharides.

Catabolic:

It is the break down of large complex molecules to the smaller molecules.

For example during cellular respiration sugar molecules break down and generate energy.

Option D=  Break down and synthesis

The break down and synthesis are also reverse pair of each others. The  synthesis involve the formation of molecules form smaller component while the break down involve destruction of molecules into smaller units.

8 0
3 years ago
A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.
kondaur [170]

The question is incomplete, complete question is:

A fluid occupying 3.2 m^3 of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = 3.2 m^3

Density of the fluid = D

D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3

Specific volume is the reciprocal of the density :

V_{specific}=\frac{1}{Density}

Specific volume of the fluid = S_v

S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg

2)

Density of the fluid in English Engineering units  = D (lb/ft^3)

1 kg = 2.20462 lb

1 m = 3.280 ft

D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3

Specific volume of the fluid :

=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb

3)

Density of the fluid in British Gravitational System units  = D (slug/ft^3)

1 kg = 0.06852 slug

1 m = 3.280 ft

D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3

Specific volume of the fluid :

=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug

7 0
3 years ago
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