2. In a chemistry experiment, 10 g of urea NH2CONH2 (s) is dissolved in 150 ml of water in a simple calorimeter. A
temperature decrease of 3.7°C is measured. Calculate the molar enthalpy of solution for urea.
​

Answer:

5m/s²

Explanation:

Given parameters:

Initial velocity  = 10m/s due east

Final velocity  = 30m/s due east

Time = 4s

Unknown:

Acceleration of the car = ?

Solution:

Acceleration is the rate of change of velocity with time.

Mathematically;

 Acceleration  = [tex]\frac{v - u}{t}[/tex]  

v is the final velocity

u is the initial velocity

t is the time taken

  Insert the parameters and solve;

  Acceleration  = [tex]\frac{30 - 10}{4}[/tex]   = 5m/s²

Answer:

241.08039999999588

Explanation:

Answer:

2.43 moles of water are required.

Explanation:

Given data:

Moles of H₂O required = ?

Moles of NO₂ present = 7.30 mol

Chemical equation:

3NO₂+ H₂O  →     NO + 2HNO₃

Now we will compare the moles of NO₂ and H₂O.

                    NO₂       :           H₂O

                      3          :              1

                   7.30        :          1/3×7.30 = 2.43 mol

Answer:

I believe it is "in industrialized areas"

Explanation:

According to Calvert Academy it says "Only 25% of the world's population lives in industrial areas. Hope this helps :)

Gold have a greater mass

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

The unit of density can be expressed in g/cm³ or kg/m³  

Density formula:  

[tex]\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}[/tex]

ρ = density  

m = mass  

v = volume  

[tex]\tt mass=\rho\times V\\\\mass=10.5\times 5=52.5~g[/tex]

[tex]\tt mass=\rho\times V\\\\mass=19.3\times 5=96.5~g[/tex]

Molecules which have a larger size as compared to pores of cell membrane then they can't pass through it.

Water, carbon dioxide, glucose and oxygen are some of the molecules that can pass through the cell membrane of the cell by the process of diffusion while on the other hand, large molecules such as sugars and proteins can't enter inside the cell due to their large size.

The cell has semipermeable membrane which has small pores. Through these pores small size materials can pass through whereas large molecules can't pass through it due to their large size so we can conclude that molecules which have a larger size as compared to pores of cell membrane then they can't pass through it.

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We know, when temperature and pressure is constant :

[tex]\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}[/tex]      ....1)

[tex]n_1 = 3\ moles,\ V_1 = 1.50\ L[/tex]

[tex]n_2 = 3 + 0.80 = 3.80 \ moles[/tex]

Let, final volume is [tex]V_2[/tex].

Putting all values in equation 1), we get :

[tex]\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}\\\\\dfrac{1.50}{3}=\dfrac{V_2}{3.80}\\\\V_2 = \dfrac{1.50\times 3.80}{3}\\\\V_2 = 1.9\ L[/tex]

Therefore, volume (in L) of the balloon if 0.80 moles of gas are added is 1.9 L.

Hence, this is the required solution.

Answer:

The specific  heat  capacity  of  unknown  metal  is  calculated  using the  following  formula

Q=MC  delta  T  where

Q= heat(862 j)

M=  mass  (17 g)

C=  specific  heat capacity=?

delta T = change in  temperature  (  35-10= 25 c)

by making  C the  subject  of the  formula  C=Q/m  delta T

=  862 j/( 17 g  x25 c) = 2.028 j/g/c

Explanation:

Answer:

The molar mass and molecular weight of Al(CH3CO2)3 is 204.1136.

Explanation:

Answer:

168.1 g/mol

General Formulas and Concepts:

Chemistry

Explanation:

Step 1: Define

Al(CH₃O₂)₃

Step 2: Find molar masses

Molar Mass of Al - 26.98 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Al(CH₃O₂)₃ - 26.98 + 3(12.01) + 9(1.01) + 6(16.00) = 168.1 g/mol

Answer:

6 mol H₂O

General Formulas and Concepts:

Chemistry

Explanation:

Step 1: Define

RxN:     2H₂ + O₂ → 2H₂O

Given:   3 moles O₂

Step 2: Stoichiometry

[tex]3 \ mol \ O_2(\frac{2 \ mol \ H_2O}{1 \ mol \ O_2} )[/tex] = 6 mol H₂O

Step 3: Check

We are given 1 sig fig.

Our final answer has 1 sig fig, so no need to round.

Answer:

Molarity = 2 M

Explanation:

Given data:

Molarity of solution = ?

Number of moles = 2 mol

Volume of water = 1 L

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula

Molarity = number of moles of solute / L of solution

Molarity = 2 mol / 1 L

Molarity = 2 M

Answer: Units. The SI derived unit of frequency is the hertz (Hz), named after the German physicist Heinrich Hertz. One hertz means that an event repeats once per second.

The frequency is the number of waves that pass a point in space during any time interval, usually one second. We measure it in units of cycles (waves) per second, or hertz. The frequency of visible light is referred to as color, and ranges from 430 trillion hertz, seen as red, to 750 trillion hertz, seen as violet.

Frequency of an electromagnetic wave is the number of wave cycles per unit time. The SI unit of frequency is Hertz or Hz. Frequency is represented by the symbol v called neu.

Waves are propagation or transfer of energy through air or vacuum. An electromagnetic wave is associated with a combined electric field and magnetic field.

Waves are associated with characteristic frequency and wavelength. Wavelength of a wave is the distance between its two consecutive crests or troughs.

Frequency of a wave is the number of wave cycles per unit time. Its is expressed in the unit of Hertz or Hz. Frequency and wavelength are in inverse relation. The shortest waves have greater energy and greater frequency.

Frequency is inverse of time of propagation. So that it can also be used in s⁻¹ which is equal to one hertz.

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Answer:

12ml should be the answer

Explanation:

Answer:

[tex]$\Delta S = -10.95 \ \frac{J}{K}$[/tex]

Explanation:

Given :

Number of mole, n = 0.720 mol

Initial volume, [tex]$V_1$[/tex] = 24. 32 L

Final volume, [tex]$V_2$[/tex] = 3.90 L

We know entropy change,

[tex]$\Delta S = c_V\ln \left(\frac{T_2}{T_1}\right) + n R \ln \left(\frac{V_2}{V_1}\right)$[/tex]

For isothermal compression temperature constant,

[tex]$\Delta S = n R \ln \left(\frac{V_2}{V_1}\right)$[/tex]

[tex]$\Delta S = 0.720 \times 8.314 \times \ln \left(\frac{3.90}{24.32}\right)$[/tex]

[tex]$\Delta S = -10.95 \ JK^{-1}$[/tex]

[tex]$\Delta S = -10.95 \ \frac{J}{K}$[/tex]

Answer:

6

8

6

Explanation:

 Isotope given:

                       ¹⁴₆C

In specie written as this;  

   Superscript  = Mass number

   Subscript  = Atomic number

To find the protons, it is the same as the atomic number;

                Protons  = Atomic number  = 6

Neutrons have no charges;

 Neutrons  = Mass number - Atomic number  =

  Neutrons  = 14 - 6  = 8

The number of electrons is the same as the atomic number = 6

Answer:

[tex]M=0.173M[/tex]

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of solute divided by the volume of the solution in liters and has units of mol/L or M:

[tex]M=\frac{n}{V}[/tex]

For the solute, Fe3Br2, which has a molar mass of 327.3430 g/mol, the moles in 42.389 g are:

[tex]n=42.389g*\frac{1mol}{327.3430g} =0.12949mol[/tex]

Thus, since the volume in liters is 0.750 L from those given 750 mL, the molarity turns out:

[tex]M=\frac{0.12949mol}{0.750L}\\\\M=0.173M[/tex]

Best regards!

Answer:

ethene

Explanation:

just took the test :D

Answer: Where is the folowing?

Explanation:

[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Kinematics.

Here, we know that, Velocity = Distance / Time,

So here, Distance = 10km = 10×1000 = 10000 metres.

, Time = 14 min 30 sec = 870 seconds,

so now, we get velocity as,

=> V = 10000 ÷ 870 => 11.49 m/s .

Hence, Velocity is 11.49 m/s.

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

[tex]2H^+ + 2e^- \to H_2 \ \ \ \mathbf{E^0_{red}= 0.00 \ V}[/tex]

At the anode:

At oxidation reaction is:

[tex]Zn \to Zn^{2+} +2e^- \ \ \ \mathbf{E^0_{ox} = 0.76 \ V}[/tex]

The overall equation for the reaction is:

[tex]\mathbf{Zn + 2H^+ \to Zn^{2+} + H_2}[/tex]

The overall cell potential is:

[tex]\mathbf{E^0_{cell}= E^0_{ox} + E^0_{red}}[/tex]

[tex]\mathbf{E^0_{cell}= 0.76 \ V +0.00 \ V}[/tex]

[tex]\mathbf{E^0_{cell}= 0.76\ V}[/tex]

Using the formula for the Nernst equation:

[tex]E = E^0 - ( \dfrac{0.0591}{n})log (Q)\\[/tex]

where;

E = 0.66

(Zn^2+)=0.22 M

Then

[tex]0.66 =0.76- ( \dfrac{0.0591}{2})log \bigg ( \dfrac{[Zn^{2+} ] PH_2}{[H^+]^2} \bigg )[/tex]

[tex]0.66 =0.76- 0.02955 * log \bigg ( \dfrac{0.22*0.87}{[H^+]^2} \bigg )[/tex]

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log  [H⁺]

4.118  = - 2  log  [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) ----> Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M =  0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 =  0.76 - 0.0592/2logQ

0.660 - 0.76  =  - 0.0592/2logQ

-0.1 =  - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 =  [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92  × 10^-3/[H^+]^2

[H^+]^2 = 7.92  × 10^-3/2.39 × 10^3

[H^+] = 1.82  × 10^-3

pH = -log[H^+]

pH = -log[ 1.82  × 10^-3]

pH = 2.7

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Answer:

Percentage of HCL = 37.11 % (Approx)

Molarity = 16.17 m (Approx)

Explanation:

Given:

Molarity = 12.0 M  of hcl

Density = 1.18 g/ml

Computation:

1L of hcl solution contains = 12 moles

Mass = Moles x Molar mass

Mass of HCl = 12 x 36.5

Mass of HCl  = 438 g  = 0.438 kg

Mass of 1 L solution =  1.18 x 1,000 ml

Mass of 1 L solution = 1180 g

Percentage of HCL = [438/1180]100

Percentage of HCL = 37.11 % (Approx)

Molarity = 12 / [1.18 - 0.438]

Molarity = 12 / [0.742]

Molarity = 16.17 m (Approx)

Concepts used:

1 mole of an element or a compound has 6.022 * 10²³ formula units

So, we can say that: Number of formula units = number of moles * 6.022*10²³

number of moles of an element or a compound = given mass/molar mass

__________________________________________________________

003 - Number of CaH₂ formula units in 6.065 grams

Number of Moles:

We know that the molar mass of CaH₂ is 42 grams/mol

Number of Moles of CaH₂ = given mass/molar mass

Number of moles = 6.065 / 42

Number of moles = 0.143 moles

Number of Formula units:

Number of formula units = number of moles * 6.022*10²³

= 0.143 * 6.022 * 10²³

= 0.86 * 10²³ formula units

__________________________________________________________

004 - Mass of 6.34 * 10²⁴ formula units of NaBF₄

Number of Moles:

We mentioned this formula before:

Number of formula units = number of moles * 6.022*10²³

Solving it for number of moles, we get:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variable

Number of moles = 6.34 * 10²⁴ / 6.022*10²³

Number of moles=  10.5 moles

Mass of 10.5 moles of NaBF₄:

Molar mass of NaBF₄ = 38 grams/mol

Mass of 10.5 moles = 10.5 * molar mass

Mass of 10.5 moles = 10.5 * 38

Mass = 399 grams

__________________________________________________________

005 - Number of moles in 9.78 * 10²¹ formula units of CeI₃

Number of Moles:

We have the formula:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variables

Number of Moles = 9.78 * 10²¹ / 6.022*10²³

Number of Moles = 1.6 / 10²

Number of Moles = 1.6 * 10⁻² moles   OR   0.016 moles

Answer:

Molecular formula of the compound = H₂CO₃

Explanation:

First, the empirical formula of the compound is determined

Percentage by mass of each element is given as shown below:

H = 3.3% ; C = 19.9%; O = 77.4%

Mole ratio of the elements= percentage mass/ molar mass

H = 3.3/ 1 = 3.3

C = 19.3/12 = 1.6

O = 77.4/16 = 4.8

whole number ratio is obtained by dividing through with the smallest ratio

H = 3.3/1.6; C = 1.6/1.6; O = 4.8/1.6

H : C : O = 2 : 1 : 3

Empirical formula = H₂CO₃

Molecular formula/mass = n(empirical formula/mass)

60 = n(2*1 + 12*1 + 16*3)

60 = n(62)

n = 60/62 = 0.96

n is approximately = 1

Therefore, molecular formula of the compound = (H₂CO₃) * 1

Molecular formula of the compound = H₂CO₃

The molecular formula of a compound composed of 3.3% H, 19.3% C, and 77.4% O and a molar mass of approximately 60 g/mol is CH2O3.

HOW TO CALCULATE MOLECULAR FORMULA:

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Answer:

the process of growth or increase, typically by the gradual accumulation of additional layers or matter.

Explanation:

Answer:

The process of growth or increase, typically by the gradual accumulation of additional layers or matter.

Explanation:

Answer:

i podwmk

Explanation: