If a rigid can of volume 10.0 L starts at a temperature of 27 oC and 800.0 torr and is heated to 327 oC, what is the new pressure in torr?

Answer:

1.25×10²⁵ molecules of ethane contains 124.56 moles of Hydrogen atoms.

Answer:

relating to or denoting the business of entertaining clients, conference delegates, or other official visitors.

The molarity of a solution containing 20moles of NaCl dissolved in 10 liters of water is 2 mol/L.

Molarity of any solution will be calculated by using the below formula:
M = n/V, where

n = moles of solute = 20 mol

V = volume of solution or solvent = 10 L

On putting values on above equation, we get

M = 20 / 10 = 2M

Hence required molarity of NaCl is 2 M.

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Answer:

Universal indicator can show us how strongly acidic or alkaline a solution is, not just that the solution is acidic or alkaline. This is measured using the pH scale , which runs from pH 0 to pH 14.

Explanation:

~Hope this helps

Answer:

during interphase the cells start to grow

cells grow, store up energy, and take in nutrients

hope it helped :)

Answer:

The ratio of the mass of a substance to its volume is called...density.

Answer:

This is because .It doesn't reacts with dilute acid .It is a coinage metal.

Answer:

Indicator?

Explanation:

not sure though

Answer:

A reduction potential measures the tendency of a molecule to be reduced by taking up new electrons. ... Standard reduction potentials can be useful in determining the directionality of a reaction. The reduction potential of a given species can be considered to be the negative of the oxidation potential.

Explanation:

Answer:

see calculations below

Explanation:

Given:     HOAc     ⇄    [H⁺]     +    [OAc⁻]            

C(i)          0.12M             0M           0.15M

mix  => 5.55ml(0.092M NaOH) / (50ml + 5.55ml)

            = 0.00555(0.092)mole NaOH / 0.0555 L Soln

            = 0.0092M in NaOH is added into the initial buffer solution

            = 0.0092M in OH⁻ (NaOH is a strong base => 100% ionized)

Rxn => Addition of 0.0092M OH⁻ will react with 0.0092M H⁺ shifting buffer            .           equilibrium to the right decreasing [HOAc] and increasing [OAc⁻] by    .           0.0092M each.

Therfore ...

Given:           HOAc      ⇄    [H⁺]     +       [OAc⁻]            

C(i)                0.12M             0M               0.15M

ΔC           - 0.0092M           +x            +0.0092M

C(f)             0.1108M             x                0.1592M  =>  New Concentrations            .                                                                                     after adding 0.0092M             .                                                                                     NaOH

Substituting new acid and ion concentrations into Ka expression ...

Ka = [H⁺][OAc⁻]/[HOAc] = (x)(0.1592M)/(0.1108M) = 1.75 x 10⁻⁵M

=> x = [H⁺](new) = (1.75 x 10⁻⁵M*)(0.1108M)/(0.1592M) = 1.22 x 10⁻⁵M in H⁺ ions

*units of Ka are Molar

FYI => Adding a strong base to a buffer solution will shift pH to more basic.

          Adding a strong acid   to a buffer solution will shift pH to more acidic.

=> (such is a good way to check that your buffer calculations are correct.)

NOTE => Question asks for moles of HOAc, Na⁺OAc⁻ & NaOH after adding base. Giving answers in terms of Molarity (moles/Liter) is same as moles. Therefore ...

[HOAc] = 0.1108M

[NaOAc] = 0.1592M

[NaOH] = ∅M (from rxn of H⁺ + OH⁻ => H₂O, all NaOH was consumed in acid/base reaction.  Remaining are only Na⁺ as a spectator ion and OH⁻ as a function of the new concentration of H⁺ => [OH⁻] = Kw/[H⁺] = 1 x 10⁻¹⁴/1.22 x 10⁻⁵ = 8.2 x 10⁻¹⁰M.

Hope this helps. :-)

Answer:

1.32 moles.

Explanation:

From the question given above, the following data were obtained:

Density of Al = 2.70 g/cm³

Volume of Al = 13.2 cm³

Number of mole of Al =.?

Next, we shall determine the mass of Al.

This can be obtained as follow:

Density of Al = 2.70 g/cm³

Volume of Al = 13.2 cm³

Mass of Al =?

Density = mass / volume

2.7 = mass of Al / 13.2

Cross multiply

Mass of Al = 2.7 × 13.2

Mass of Al = 35.64 g

Finally, we shall determine the number of mole of Al. This can be obtained as follow:

Mass of Al = 35.64 g

Molar mass of Al = 27 g/mol

Number of mole of Al =?

Mole = mass / molar mass

Number of mole of Al = 35.64 / 27

Number of mole of Al = 1.32 moles

Thus, 1.32 moles of aluminum are present in the block of the metal.

C

Tarnishing of silver is caused by hydrogen sulfide and other sulfur containing compounds in the air reacting to the silver and resulting in a fine layer of silver sulfide. The other answers don't make sense

Answer:

D

Explanation:

Answer:

1.00 moles of CO2 is the final yield.

Explanation:

Based on the reaction:

2C2H2(g)+ 5O2(g) → 4CO2(g) + 2H2O(g)

When 2 moles of acetylene (C2H2) completely reacts with oxygen, 4 moles of CO2 are produced.

To solve this question we must use the chemical equation knowing: 2 moles C2H2 = 4 mol CO2

When 0.500 moles of acetylene react:

0.500 moles C2H2 * (4mol CO2 / 2mol C2H2) =

There is an increase in entropy because the energetically equivalent ways to arrange the components of the  system increased. Option A and C.

The term entropy refers to the degree of disorder in a system. Thus we know that any change that leads to an increase in the number of particles will lead to an increase in entropy.

Hence, we can see from the diagram that the energetically equivalent ways to arrange the components of the  system increased hence there is a positive increase in entropy. The correct options are therefore A and C.

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The mass of the kool aid needed to prepare the solution is 81.2 g.

The term concentration has to do with the amount of substance present in solution. We are told to make calculations regarding the preparation of a 237.5 ml kool-aid solution with a concentration of 1M.

Litres of water needed = 237.5/1000 = 0.2375 L

Molarity of kool-aid needed = 1 M

number of moles = 1 M *  0.2375 L = 0.2375 moles

0.2375 moles = mass of kool aid/molar mass

0.2375 moles = mass/ 342 g/mol

mass = 0.2375 moles * 342 g/mol

mass = 81.2 g

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156.975kj heat is required to raise the temperature of 0.50 kg of liquid water from 15℃ to 90℃.

The term specific heat is defined as the quantity of heat required to increase the temperature of one gram of a substance by one Celsius degree.

To calculate the energy required to raise the temperature of any given substance,

Given:

The mass of the material, m

The temperature change that occurs, ΔT

The specific heat capacity of the material, c (which you can look up). This is the amount of heat required to raise 1 gram of that substance by 1°C.

Q = m × c × ΔT

For water, the value of c is 4.186 J/g°C

Therefore,

Q = 500 × 4.186 × 75

= 156,975j

= 156.975kj

Thus, 156.975kj heat is required to raise the temperature of 0.50 kg of liquid water from 15℃ to 90℃.

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Answer: a

Explanation:

this is common sense

Answer:

C'est fini sous la lettre A. Ice crystals from forming inside it's body

Answer:

acceleration

pls dont report me

Answer:

[tex]161.9gZn[/tex]

Explanation:

Hello there!

In this case, according the given balanced chemical reaction:

[tex]Zn + 2HCl\rightarrow H_2 + ZnCl_2[/tex]

It is possible to evidence the 1:1 mole ratio of hydrogen (molar mass = 2.02 g/mol) to zinc (atomic mass = 65.41 g/mol) which is used to calculate the grams of the latter needed for the production of 5 grams of the former via stoichiometry:

[tex]5gH_2*\frac{1molH_2}{2.02gH_2}*\frac{1molZn}{1molH_2} *\frac{65.41gZn}{1molZn}\\\\=161.9gZn[/tex]

Best regards!

Explanation:

T=?

Given

n=3.33×10³

V=478mL=4.78L

P=750torr= 0.987 Atm

simply use the formula

PV =nRT

where R is universal gas constant

0.987×4.78= 3.33×10³×8.314×T

T = 0.000170°c

T= 273.15 K!

The absolute temperature of the gas at which 3.33x10³ mol occupies 478 mL at 750 torr is 1.726 × 10-³K

HOW TO CALCULATE TEMPERATURE:

PV = nRT

Where;

According to this question,

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Answer:

3.089 L

Explanation:

From the given information, provided that the no of moles and the temperature remains constant;

[tex]P_1[/tex] = 15.6 psi

[tex]V_1[/tex] = ???

[tex]P_2[/tex] = 25.43 psi

[tex]V_2[/tex] = 1.895 L

Using Boyle's law:

[tex]P_1V_1 =P_2V_2 \\ \\ V_1 = \dfrac{P_2V_2}{P_1} \\ \\ V_1 = \dfrac{25.43 \times 1.895}{15.6} \\ \\ \mathbf{ V_1 = 3.089 \ L}[/tex]

Answer:

0.559 L

Explanation:

Step 1: Given data

Moles of sodium iodide (n): 0.405 mol

Molar concentration of sodium iodide (M): 0.724 M (0.724 mol/L)

Step 2: Calculate the volume of solution (V)

The molarity is equal to the moles of solute divided by the liters of solution.

M = n/V  

V = n/M  

V = 0.405 mol/(0.724 mol/L) = 0.559 L

(hope this helps can i plz have brainlist :D hehe)

Answer:

Bleach, seawater, blood, water, coffee, orange, lemon

Explanation:

The pH is a measurement in chemistry defined as the -log [H⁺]. pH is used in quality control of materials. The pH is a scale from 1 (Acids) to 14 (Bases). The pH of 7 is for neutral solutions

Based on the information of the image, the most basic is bleach, then the seawater, the blood, water (Neutral), coffee, orange and, the most acidic is lemon.

The order is:

Explanation:

The two main categories of energy are potential energy (stored

energy) and kinetic energy (energy presently doing

work or causing change)...I hope this helps.

Answer:

Velocity

Explanation:

Synonyms are words/phrases that mean the exact or nearly exact same in meaning. Velocity, pace, and speed all mean close to the same thing as far as the rate in which something is moving.

Answer:

2.75 × 10⁻⁶ M/s

1.69 × 10⁻⁶ M/s

9.23 × 10⁻⁻⁷ M/s

4.43 × 10⁻⁻⁷ M/s

2.1 × 10⁻⁻⁷ M/s

Explanation:

We have the following information for the isomerization of methyl isonitrile

Time (s)      [CH₃NC] (M)

  0              0.0165

2000          0.0110

5000          0.00591

8000          0.00314

12000         0.00137

15000         0.00074

To calculate the average rate of reaction (r) for each interval, we need to use the following expression:

r = -Δ[CH₃NC]/Δt

Interval 0-2000 s

r = - (0.0110 M-0.0165 M)/2000 s - 0 s = 2.75 × 10⁻⁶ M/s

Interval 2000-5000 s

r = - (0.00591 M-0.0110 M)/5000 s - 2000 s = 1.69 × 10⁻⁶ M/s

Interval 5000-8000 s

r = - (0.00314 M-0.00591 M)/8000 s - 5000 s = 9.23 × 10⁻⁻⁷ M/s

Interval 8000-12000 s

r = - (0.00137 M - 0.00314 M)/12000 s - 8000 s = 4.43 × 10⁻⁻⁷ M/s

Interval 12000-15000 s

r = - (0.00074 M - 0.00137 M)/15000 s - 12000 s = 2.1 × 10⁻⁻⁷ M/s

The average rates of reactions for the time interval between each measurement are

2.75×10⁻⁶ M/s

1.70×10⁻⁶ M/s

9.23×10⁻⁷ M/s

4.43×10⁻⁷ M/s

2.10×10⁻⁷ M/s

From the question,

We are to calculate the average rate of reaction for each of the measurement.

First, we will create a proper table for the data

Time (s)       [CH3NC] (M)

0                    0.0165    

2,000            0.0110

5,000            0.00591      

8,000            0.00314      

12,000          0.00137

15,000          0.00074

The average rate of reaction is given by the formula

[tex]Average\ rate\ of\ reaction = \frac{Final\ concentration - Initial\ concentration}{Final\ time - Initial\ time}[/tex]

1.

[tex]Average\ rate\ of\ reaction = \frac{0.0110-0.0165}{2000 -0 }[/tex]

Average rate of reaction = 2.75×10⁻⁶ M/s

2.

[tex]Average\ rate\ of\ reaction = \frac{0.00591-0.0110}{5000 -2000 }[/tex]

Average rate of reaction = 1.70×10⁻⁶ M/s

3.

[tex]Average\ rate\ of\ reaction = \frac{0.00314-0.00591}{8000 -5000 }[/tex]

Average rate of reaction = 9.23×10⁻⁷ M/s

4.

[tex]Average\ rate\ of\ reaction = \frac{0.00137-0.00314}{12000 -8000 }[/tex]

Average rate of reaction = 4.43×10⁻⁷ M/s

5.

[tex]Average\ rate\ of\ reaction = \frac{0.00074-0.00137}{15000 -12000 }[/tex]

Average rate of reaction = 2.10×10⁻⁷ M/s

Hence, the average rates of reactions for the time interval between each measurement are

2.75×10⁻⁶ M/s

1.70×10⁻⁶ M/s

9.23×10⁻⁷ M/s

4.43×10⁻⁷ M/s

2.10×10⁻⁷ M/s

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Answer:

3.1 * 10^-14

Explanation:

Note that E°cell = 0.0592/n log K

We can obtain E°cell from the standard reduction potentials of cadmium and hydrogen

Anode reaction

H2(g) ----> 2H+ + 2e

Cathode reaction

Cd^2+(aq) + 2e -----> Cd(s)

E°cell = E°cathode - E°anode

E°cathode = –0.40 V

E°anode = 0 V

E°cell =  –0.40 V - 0 V

E°cell = –0.40 V

E°cell = 0.0592/n log K

Where n=2 electrons transferred

–0.40 = 0.0592/2 log K

–0.40 = 0.0296 log K

log K = –0.40/0.0296

log K = -13.5135

K = Antilog ( -13.5135)

K = 3.1 * 10^-14