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3 years ago

How can you show yourself compassion?

Chemistry

2 answers:

RoseWind [281]3 years ago

6 0

Answer:

The definition of compassion is "sympathetic pity and concern for the sufferings or misfortunes of others." so I guess show that to yourself

Explanation:

examples are eating, cutting yourself a break on something difficult, drinking water and taking care of yourself going easy on your body.

SSSSS [86.1K]3 years ago

3 0

Answer:

eat healthy, exercise, be kind to yourself, things like that

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Within the context of a person trying to explain why she did well on a psychology exam, an internal attribution is to an externa

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Instinct to learning.

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4 years ago

When Thomas Edison first sold electricity, he used zinc coulometers to measure charge consumption. (In a coulometer the cathode

IrinaK [193]

Answer:

The answer is " $42537.93 \ C$ ".

Explanation:

Molar mass : $Z_n = 65.38 \ \frac{g}{mol}\\\\$

$\to Zn^{2+} + 2e^{-} = Zn\\\\$

$\therefore\\\\ \to 65.38 \ g \ of\ Z_n \ require: \\\\ \to 2 \times 96500\ C\\\\\to 14.41 \ g\ require,\\\\ \to \frac{(14.41 \times 2 \times 96500)}{65.38} \ \frac{g}{mol} = 42537.93 \ C$

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3 years ago

Compare the properties of protons, neutrons and electrons

sergejj [24]

Answer:

Electrons are a type of subatomic particle with a negative charge. ... Protons and neutrons have approximately the same mass, but they are both much more massive than electrons (approximately 2,000 times as massive as an electron). The positive charge on a proton is equal in magnitude to the negative charge on an electron

Explanation:

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3 years ago

Read 2 more answers

A gas mixture being used to simulate the atmosphere of another planet at 23°c consists of 337 mg of methane, 148 mg of argon, an

Karolina [17]

The total pressure of the mixture is 65.5 kPa.

According to Dalton's Law of Partial Pressure,

The partial pressure of gas = Mole fraction of gas × Total pressure

Total Pressure = Sum of all the gases partial pressures

The number of moles of methane is,

$Moles \: of \: methane \: (16 g/mol) = 337 \: mg \times \frac{1 g}{1000 mg} \times \frac{ 1 mol}{16 g }$

= 0.021 mols

The moles of methane are 0.021 mols.

The number of moles of the argon,

$Moles \: of \: argon (40 g/mol) = 148 \: mg \times \frac{ 1 g}{1000 mg } \times \frac{ 1 mol}{40 g}$

= 0.003 mols

The number of moles of argon is 0.003 mols.

The number of moles of nitrogen is,

$Moles \: of \: nitrogen (28 g/mol) = 296 \: mg \times \frac{ 1 g}{1000 mg} \times \frac{ 1 mol/}{28 g}$

= 0.010 mols

The number of moles of nitrogen is 0.010 mols.

The total number of moles is,

= 0.021 + 0.003 + 0.010

= 0.034 mols

$Mole \: fraction = \frac{ Moles \: of \: solute }{Total \: number \: of \: moles \: of \: soulte \: and \: solvent}$

$= \frac{ 0.010 }{ 0.034}$

= 0.29

$0.29 \: P _{total} = 19 \: kPa$

$P _{total} = \frac{ 19 \: kPa }{0.29}$

= 65.5 kPa

Therefore, the total pressure of the mixture is 65.5 kPa.

To know more about Dalton's law, refer to the below link:

brainly.com/question/14119417

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6 0

1 year ago

The rate of formation of C in the reaction 2 A + B →2 C + 3 D is 2.7 mol dm−3 s −1 . State the reaction rate, and the rates of f

Zielflug [23.3K]

Answer:

Rate of reaction = $1.35mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $4.15mol.dm^{-3}.s^{-1}$

Explanation:

According to laws of mass action for the given reaction,

$Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}$

where, $-\frac{\Delta [A]}{\Delta t}$ is rate of consumption of A, $-\frac{\Delta [B]}{\Delta t}$ is rate of consumption of B, $\frac{\Delta [C]}{\Delta t}$ is rate of formation of C and $\frac{\Delta [D]}{\Delta t}$ is rate of formation of D

Here $\frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}$

So, Rate of reaction = $(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $(\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $(\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $(\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$

3 0

3 years ago