at a certain temperature nitrogen dioxide gives dinitrogen tetroxide in a second order reaction. the rate constant for this reaction is 4.50/M min. what will be the concentration of nitrogen dioxide after 0.355 min, if one starts with an initial concentration of 0.281 M​

Answer:

-0.276 Degrees C

Explanation:

kf of water is 1.86

Freezing point depression= m x kf x i

i= ions present ( K+ Cl-) 1=2

1.86x2X.743= .276

Since its freezinf point depression the freezing point will lower

so

0-.276= -.276

Answer: -2.764

Explanation:

Answer:

The decay takes 6min

Explanation:

The decay of an isotope as Thallium-208 follows the equation:

ln[A] = -kt + ln[A]₀

Where [A] is amount of isotope after time t

k is decay constant = ln 2 / t(1/2)

[A]₀ is initial amount of the isotope

Replacing:

[A] = 15.0g

k = ln 2 / t(1/2) = ln 2 / 3min = 0.23105 min⁻¹

t = ?

[A]₀ = 60.0g

ln[A] = -kt + ln[A]₀

ln[15.0g] = -0.23105 min⁻¹*t + ln[60.0g]

-1.38629 = -0.23105 min⁻¹*t

6min = t

Answer:

UK stop spending a lot on clothes if u don't wear half of them

Answer:

No

Explanation:

Rust is an iron oxide, a usually reddish-brown oxide formed by the reaction of iron and oxygen in the catalytic presence of water or air moisture.

Answer:

96.92 K

Explanation:

Applying,

V/T = V'/T'..................... Equation 1

Where V = Initial Volume, T = Initial Temperature, V' = Final Volume, T' = Final Temperature.

Make T the subject of the equation

T = VT'/V'................. Equation 2

From the question,

Given: V = 268 mL, V' = 824 mL, T' = 25°C = (25+273) = 298 K

Substitute these values into equation 2

T = (268×298)/824

T = 96.92 K

Hence the Initial temperature of Nitrogen gas is 96.92 K

Answer:

thank uuuuuuu for freeeee pointssssss!!!!!..

Explanation:

hope it is helpful to you

Answer: no thank u

Explanation:

Answer: no

Explanation:

Answer:

wavelength

Explanation:

it is the distance between two crests ot two troughs

Answer:

14.4g

Explanation:

Answer:

Beta emission

Explanation:

In beta emission, a neutron is converted into a proton thereby emitting an electron and a neutrino. A neutrino is a particle that serves to balance the spins.

When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.

Hence, in beta emission, the daughter nucleus is found one pace to the right of the parent in the periodic table.

Answer:

A. Mass

Explanation:

If it takes up space it has to have mass. Not everything that is matter has a color, certain length or width.

Answer: 114.22852 grams

HOPE THIS HELPS

Explanation:

HCL you can do it yourself .try again

I know number 1 is 58.04 g/mol

Solution :

Compound                      Ksp

[tex]$PbF_2$[/tex]                               [tex]$3.3 \times 10^{-8}$[/tex]

[tex]$Ni(CN)_2$[/tex]                        [tex]$3 \times 10^{-23}$[/tex]

FeS                                [tex]$8 \times 10^{-19}$[/tex]

[tex]$CaSO_4$[/tex]                           [tex]$4.93 \times 10^{-5}$[/tex]

[tex]$Mg(OH)_2$[/tex]                      [tex]$5.61 \times 10^{-12}$[/tex]

Ksp of [tex]$Ni(CN)_2 << Ksp \text{ of}\ \ Mg(OH)_2$[/tex] and both compounds dissociate the same way. Hence [tex]$Mg(OH)_2$[/tex] is more soluble than [tex]$(B). \ Ni(CN)_2$[/tex]

[tex]$Mg(OH)_2$[/tex]  is less soluble than [tex]$(A). \ \ PbF_2 \ ()Ksp \ PbF_2 > Ksp \ \text{ of } \ Mg(OH)_2$[/tex]

It is not possible to determine CD - [tex]$FeS \text{ or} \ CaSO_4$[/tex]  is more or less soluble than [tex]$Mg(OH)_2$[/tex]  as though they have a different Ksp values their molecular dissociation is also different and they may have a close solubility values.

[tex]$Ni(OH)_2$[/tex]  can be directly compared with PbS, [tex]$AlPO_4, MnS$[/tex]

[tex]$\text{For } \ Ni(OH)_2$[/tex]

[tex]$AB_2(s) \rightarrow A^{2+} + 2B^{-}$[/tex]

[tex]$Ni(OH)_2(s) \rightarrow Ni^{2+} + 2OH^-$[/tex]

100

1-s s 2s

Ksp = [tex][A2+][B-]^2 = s \times (2s)^2 = 4s^3[/tex]

Hence they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

For Silver Chloride

[tex]$AB(s) \rightarrow A^{x+}+B^{x-}$[/tex]

[tex]$AgCl(s) \rightarrow Ag^+ + Cl^-$[/tex]

1 0 0

1 - s s s

Ksp [tex]$=[A^{x+}][B^{x-}]=s \times s = s^2$[/tex]

Hence, they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

Answer:

i gotchu

Explanation:

its the 3rd raikages will eassyyy

Answer:

mass of chlorine = 10.4 grams

Explanation:

Mass of the solution = 5.20 × 10⁶ g

ppm chlorine by mass = 2.00

Recall that:

[tex]ppm = \dfrac{mass \ of \ chorine \ gas (solute)}{ mass of the solution }\times 10^6[/tex]

[tex]2 = \dfrac{mass \ of \ chorine \ gas (solute)}{ 5.20\times 10^6}\times 10^6[/tex]

mass of chlorine = 5.20 × 2.0

mass of chlorine = 10.4 grams

Answer: The element B will have ONE unpaired electron in the p orbital.

Explanation:

The electronic configuration of each given element is as follows.

Atomic number of calcium (Ca) is 20.

Ca: [tex]1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}[/tex]

Atomic number of nitrogen (N) is 7.

N: [tex]1s^{2} 2s^{2} 2p^{3}[/tex]

Atomic number of boron (B) is 5.

B: [tex]1s^{2} 2s^{2} 2p^{1}[/tex]

Atomic number of argon (Ar) is 18.

Ar: [tex]1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}[/tex]

Atomic number of bromine (Br) is 35.

Br: [tex][Ar] 4s^{2} 3d^{10} 4p^{5}[/tex]

Therefore, boron is the only element that have one unpaired electron in the p-orbital.

Thus, we can conclude that element B will have ONE unpaired electron in the p orbital.

Answer:

%m/m = 0.9975%

Xₐ = 0.0039

Explanation:

In order to do this, we need various data. First to all, we need tje molecular mass of the ethanol. this can be obtained in handbooks, or simply taking the atomic weights of carbon (12 g/mol), Hydrogen (1 g/mol) and oxygen (16 g/mol) and summing those values:

MM C₂H₆O = (2*12) + (6*1) + (16*1) = 46 g/mol

Now, there is an expression that is commonly used to determine the molarity of a solution given the mass percent and density, and assuming of course, 1 liter of solution:

M = d * %m/m * 1000 / MM * 100     (1)

From here, we can solve for %m/m:

%m/m = M * MM * 100 / d * 1000

As the problem is not saying the volume of solution, we can easily assume we have 1 liter of solution. Therefore, the %m/m replacing the given data is:

%m/m = 0.216 *46 * 100 / 0.996 * 1000

To get the mole fraction, we first need to get the volume of solvent. From the density, we can get the mass of solution:

m = V * d

m = 1000 * 0.996 = 996 g of solution.

The mass of solute is:

m = 0.216 mol/L * 46 g/mol

m = 9.936 g/L, or simply we have 9.936 g of ethanol in 1 L of solution.

The mass of solvent:

solvent = 996 - 9.936 = 986.064 g

The molecular mass of water, so we can get the moles is 18 g/mol so:

moles water = 986.064 / 18 = 54.78

Finally the mole fraction:

Xₐ = 0.216 / (0.216 + 54.78)

Hope this helps

Answer

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer:

See Explanation

Explanation:

[tex]\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n[/tex]

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) [tex]\frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy[/tex]

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J