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2 years ago

PLEASE HELP TEST does anyone know the answer???? density question

Chemistry

2 answers:

CaHeK987 [17]2 years ago

8 0

Answer:

The answer is honey.

Explanation:

DerKrebs [107]2 years ago

6 0

Answer: alcohol

Explanation:

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Organic acids and bases in their ___________ form are soluble in organic solvents, but the corresponding _________ are more wate

gizmo_the_mogwai [7]

Answer:

a. neutral

b. salts

c. salt

Explanation:

Organic salts are a dense number of ionic compounds with innumerable characteristics. They are previously derived from an organic compound, which has undergone a transformation that allows it to be a carrier of a charge, and that in addition, its chemical identity depends on the associated ion.

Organic salts are usually stronger acids or bases than inorganic salts. This is because, for example, in the amine salts, it has a positive charge due to its bond with an additional hydrogen: A + -H. Then, in contact with a base, donate the proton to be a neutral compound again

RA + H + B => RA + HB

H belongs to A, but it is written as it is involved in the neutralization reaction.

On the other hand, RA + can be a large molecule, unable to form solids with a crystalline network stable enough with the hydroxyl anion or oxyhydrile OH–.

When this is so, salt RA + OH– behaves as a strong base; even as basic as NaOH or KOH

3 0

2 years ago

pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha

Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

* $W=1142.86Joule$

* $Q=997.7J$

* $H=2140.5J$

Explanation:

From the question we are told that:

Temperature $T=337K$

Pressure $P=(60-55)Pa*10^5$

Volume $V=(1.6-1.4)m^3*10^{-3}$

Generally the equation for gas Constant is mathematically given by

$\frac{P_2}{P_1}=\frac{V_1}{V_2}^n$

$\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n$

$n=0.65$

Therefore

Work-done

$W=\int{pdv}$

$W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}$

$W=1142.86Joule$

Generally the equation for internal energy is mathematically given by

$Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}$

$Q=997.7J$

Therefore

$H=Q+W$

$H=997.7J-11.42.9$

$H=2140.5J$

5 0

2 years ago

Members of a species can mate with each other and produce

Georgia [21]

Fertile offspring mate is a form of reproduction

6 0

2 years ago

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

2 years ago

What, according to you, are the necessary activities that a penetration tester needs to perform while conducting a penetration t

bogdanovich [222]

Answer: First of all planning is done to set the goals to how to test the target. then scanning is done. after that target is uncovered to see how the target exploits by the use of web application attacks. then analysis is done an tests of the penetration tests are compiled.

Explanation:

6 0

2 years ago