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2 years ago

Which of the following could be absent from the grief process

Chemistry

1 answer:

grin007 [14]2 years ago

4 0

1. C.
2. A.
3. A. or B.
4. D.
5. B.
6. B.

Sorry if they are wrong. I tried my best.

You might be interested in

How is 8.2 x 10^4 - 6.3 x 10^3 written in scientific notation

ser-zykov [4K]

Answer:

= 7.57 × 104

(scientific notation)

= 7.57e4

(scientific e notation)

= 75.7 × 103

(engineering notation)

(thousand; prefix kilo- (k))

Explanation:

Just in case this is all of them

6 0

2 years ago

Identify the units used to measure volume mass.

Strike441 [17]

Answer:

The metric system uses units such as meter, liter, and gram to measure length, liquid volume, and mass, just as the U.S. customary system uses feet, quarts, and ounces to measure these.

Volume: 1 liter is a little more than 1 quart

Mass: 1 kilogram is a little more than 2 pounds

Length: 1 centimeter is a little less than half an

Explanation:

3 0

1 year ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

2 years ago

A 2. 75-l container filled with co2 gas at 25°c and 225 kpa pressure springs a leak. When the container is re-sealed, the pressu

Serjik [45]

The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.

<h3>What is Ideal law ?</h3>

According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.

PV = nRT.

Where,

p = pressure
V = volume (1.75 L = 1.75 x 10⁻³ m³)
T = absolute temperature
n = number of moles
R = gas constant, 8.314 J*(mol-K)

Therefore, the number of moles is

n = PV / RT

State 1 :

T₁ = (25⁰ C = 25+273 = 298 K)
p₁ = 225 kPa = 225 x 10³ N/m²

State 2 :

T₂ = 10 C = 283 K
p₂ = 185 kPa = 185 x 10³ N/m²

The loss in moles of gas from state 1 to state 2 is

Δn = V/R (P₁/T₁ - P₂/T₂ )

V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N

p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)

p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)

Therefore,

Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))

= 0.0213 mol

Hence, The number of moles of gas lost is 0.0213 mol.

Learn more about ideal gas here ;

https://brainly.in/question/641453

#SPJ1

3 0

1 year ago

Based on the position of the elements in the periodic table, what is the charge of the ions they are most likely to form?

Rainbow [258]

Explanation:

The charge on an ion denotes the amount of electrons lost or gained or even shared by an atom.

An atom will lose, gain or share an equal amount of electrons that will make it stable and achieve an octet and perfect configuration.

This is often synonymous with the component number of electrons in their outermost shell. The valence shell.

For metals on the periodic table, they are always willing to give up electrons due to their large electropositivity.

Metals will give up the number of electrons in their outermost shell to become stable.

Groups 1 and Group 2 will have charges +1 and +2 on them.

For non-metals , they will gain the exact number of electrons that will make them stable. We must note that half -filled and fully filled orbitals are equally stable.

Elements in groups 6 and 7 are electronegative and will have a charge of -2 and -1 respectively.

For those in groups 3, 4 and 5 they either gain, lose or share commensurate amount of electrons that will make them stable.

Group 8 elements are stable and have no charges.

Generally on the periodic table, metals are to the left and are always positively charged. Non - metals are to the right and are negatively charged.

learn more:

Periodic table brainly.com/question/2690837

#learnwithBrainly

4 0

2 years ago