draw all the possible isomers of octane​

Answer:

The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.

Explanation:

The activity of P-32 can be calculated with the following equation:

[tex] A = \lambda N [/tex]   (1)

Where:

N: is the number of atoms of P-32

λ: is the decay constant

We can find the number of atoms of P-32 as follows:

[tex] N = \frac{N_{A}*m}{M} [/tex]  (2)

Where:

[tex]N_{A}[/tex]: is the Avogadro's number = 6.022x10²³ atoms/mol

m: is the mass of P-32 = 3.5x10⁻³ g

M: is the molar mass of the radionuclide (P-32) = 32 g/mol    

Now, the decay constant is given by:

[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]   (3)

Where:

[tex]{t_{1/2}} [/tex]: is the half-life of P-32 = 14.3 days

Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):

[tex] A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s [/tex]      

Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:

[tex] A = 3.7 \cdot 10^{13} Bq [/tex]

And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:

[tex] A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci [/tex]

Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.  

I hope it helps you!

Answer:

4 boiling point elevation

The question is incomplete, the complete question is shown in the image attached to this answer.

Answer:

a) Br^-

b) FeCl3

c) slower

d) see the first attached image

Explanation:

Aromatic compounds undergo electrophilic substitution sections in the presence of the appropriate electrophile.

In the reaction above, the Br^- nucleophile attacks the Lewis acid FeCl3. Recall that the nitro group is meta directing hence the incoming Br^+ electrophile is directed towards the meta position as shown in the image attached.

Note that the nitro group deactivates the ring towards electrophilic substitution hence the reaction is slower with nitrobenzene than with unsubstituted benzene.

Answer:

(a) m = 0.327 m.

(b) m = 4.57 m.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly considering the fact that the molality is computed by dividing the moles of solute by the kilograms of solvent, in this case water; in such a way, we proceed as follows:

(a) We firstly calculate the moles of 36.2 grams of sucrose as its molar mass is 342.3 g/mol:

[tex]\frac{36.2g}{342.3g/mol} =0.106mol[/tex]

Next, the kilograms of water in this case are 0.323 kg so that the molality will be:

[tex]m=\frac{0.106mol}{0.323kg}\\\\m=0.327m[/tex]

(b) In this case, we directly realize that the kilograms of water are now 1.889 kg so that the molality will be:

[tex]m=\frac{8.63mol}{1.889kg}=4.57m[/tex]

Clearly, the both of them in molal, m, units.

Regards!

Answer:

kjajjahahayq :/

Explanation:

a sbywsbgv usnwbhx hg xw nx hb gs

Answer:

Answer:

3,29L

Explanation:

3.29L = V2

Formula: V1/T1 = V2/T2

--------------------

Given:

V1 = 3.0 L V2 = ?

T1 = 310 K T2 = 340 K

--------------------

Plugin:

(X stands in place of V2 just to make it easier to look at)

[3.0L / 310K = X / 340K]

(3.0L / 310K = 0.01L/K)

0.01L/K = X / 340K

(multiply 340K on both sides, it cancels out on the right)

0.01L/K * 340K = X

(0.01L/K * 340K = 3.29L)

**3.29L = X**

[or]

**3.29L = V2**

Answer:

Age of rock = 6.12 × 10³ years

Note: The question is incomplete.A similar but complete question is given below.

The half-life for the radioactive decay of carbon-14 to nitrogen-14 is 5.73 x 10^3 years. Suppose nuclear chemical analysis shows that there is 0.523mmol of nitrogen-14 for every 1.000 mmol of carbon-14 in a certain sample of rock.

Calculate the age of the rock. Round your answer to 2 significant digits.

Explanation:

The half-life of a radioactive material is the time taken for half the atoms in the atomic nucleus of a material to disintegrate.

The half-life for the radioactive decay of carbon-14 to nitrogen-14 is given as 5.73 x 10³ years. This means that given 1 mole of carbon-14 is present initially, after one half-life, 0.5 moles of carbon-14 would remain.

Number of millimoles of carbon-14 remaining = 1 - 0.523 = 0.477 mmol

Number of half-lives that the carbon-14 has undergone is determined as follows:

Amount remaining = (1/2)ⁿ

where nnis number of half-lives

0.5 mmol = one half-life

0.5 = (1/2)¹

O.477 = (1/2)ⁿ = (0.5)ⁿ

㏒₀.₅(0.477) = n

n = ㏒(0.477)/㏒(0.5)

n = 1.067938829

Age of the rock = number of half-lives × half-life

Age of rock = 1.067938829 × 5.73 × 10³ years

Age of rock = 6.12 × 10³ years

Answer:

0.074 moles

Explanation:

For every mole (of any element), there are 6.022 x 10^23 atoms.

There are 4.45 x 10^22 atoms of iron.

To find the moles we divide the number of atoms by Avogadro's number

4.45 x 10^22 / 6.022 x 10^23 = 0.0738957

Don't forget sig figs

Answer:

[tex]\boxed {\boxed {\sf 0.0739 \ mol \ Fe}}[/tex]

Explanation:

We are asked to convert a number of iron atoms to moles of iron.  

We will use Avogadro's Number for this, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this problem, the particles are atoms of iron. There are 6.022 ×10²³ atoms of iron in 1 mole of iron.  

We will also use dimensional analysis to solve this problem. To do this, we use ratios. Set up a ratio using the underlined information.

[tex]\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]

Since we are converting 4.45 × 10²² atoms of iron to moles, we multiply the ratio by that value.

[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]

Flip the ratio. The value is the same, but it allows us to cancel the units of atoms of iron.

[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {1 \ mol \ Fe}{6.022 \times 10^{23} \ atoms \ Fe}[/tex]

[tex]4.45 \ \times 10^{22} *\frac {1 \ mol \ Fe}{6.022 \times 10^{23}}[/tex]

Condense into 1 fraction.

[tex]\frac {4.45 \ \times 10^{22} }{6.022 \times 10^{23}} \ mol \ Fe[/tex]

[tex]0.07389571571 \ mol \ Fe[/tex]

The original measurement of atoms ( 4.45 × 10²²) has 3 significaint figures, so our answer must have the same. For the number we calculated, that is the ten-thousandths place. The 9 to the right of this place (0.07389571571) tells us to round the 8 up to a 9.

[tex]0.0739 \ mol \ Fe[/tex]

There are approximately 0.0739 moles of iron in 4.45 × 10²² atoms of iron.  

3,3-dimethylhexane is the nomenclature of the compound

Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.

Explanation:

Plato

Answer:

We need chemistry in nursing because it deals with various kinds of drugs and the reactions of these drugs on the human body as well as with each other.

Answer:

lower the activation energy needed to start the reaction.

Explanation:

The activation energy is defined as the energy barrier that stands between reactants and products.

An enzyme is a biological catalyst. Catalysts are known to lower the activation energy of a reaction.

Hence, a catalyst lowers the activation energy of the reaction. The lower the activation energy of a reaction, the faster the reaction is expected to be.

Explanation:

here's the answer to your question

Answer:

The front passenger airbag has a volume of around 140 l and fully inflates in around 35 ms. The process is similar for side airbags (thorax airbags).

Answer:

See explanation and image attached

Explanation:

The reaction of (R)-2-chloro-3-methylbutane with potassium tert-butoxide yields a monosubstituted alkene .

Since the base is bulky, the Hoffman product predominates because attack occurs at the less hindered carbon atom to yield the major product as shown.

The alkene reacts with HBr at the secondary carbon atom to yield a carbocation intermediate which is flat and planar. Attack on either face of the carbocation yields a racemic mixture of the (2R) and (2S) products.

Rearrangement of the carbocation to yield a tertiary carbocation gives the 2-bromo-2-methyl butane product as shown in the image attached.

Answer:

Modern windshields are generally made of laminated safety glass, a type of treated glass, which consists of, typically, two curved sheets of glass with a plastic layer laminated between them for safety, and bonded into the window frame.

Explanation:

mark me brain liest if my answer is correct

Answer:

Correct ratio of carbon to hydrogen is 2:1:1

Answer:

Its actually 1:2:1

Explanation:

The molecular formula is C6H12O6 because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH2O.

Answer:

Solids are made up of cation-anion pairs, but plasmas are not

Explanation:

Answer:

Where the products are H2O and Ba(NO3)2

Explanation:

A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called neutralization reaction.

The balanced reaction is:

Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2

Where the products are H2O and Ba(NO3)2

Explanation:

here's the answer to your question

Answer:

0.8017

Explanation:

Find the molar Mass of KF

K = 39

F = 19

Total = 58

Note: these numbers are approximate. Use your periodic table to get the exact numbers.

mols = given mass / molar mass

given mass = 46.5

molar mass = 58

mols = 46.5 / 58

mols = 0.8017

Answer:

[tex]\%diff=24.0\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set up the van der Waals' equation as shown below:

[tex]p=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]

Thus, we secondly calculate the molar volume as:

[tex]v=\frac{0.8311L}{9.998mol} =0.083L/mol[/tex]

Then, we plug in the entire variables in the vdW equation to get such pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol-0.03219L/mol}-\frac{1.345L*atm/mol}{(0.08313L/mol)^2}\\\\p=615.2atm[/tex]

And the ideal gas pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol}\\\\p=496.2atm[/tex]

Finally, the percent difference:

[tex]\%diff=\frac{|496.2atm-615.2atm|}{496.2atm} *100\%\\\\\%diff=24.0\%[/tex]

Regards!

Answer:

Sr is the more metallic element

Bi is the more metallic element

O is the more metallic element

As is the more metallic element

Explanation:

One thing should be clear; metallic character increases down the group but decreases across the period.

Hence, as we move across the period, elements become less metallic. As we move down the group elements become more metallic.

This is the basis upon which decisions were made about the metallic character of each of the elements listed above.

Answer:B) 1s 2s 2p 3s 3p 4s 3d

Explanation:

The ground state electron configuration shows how the electrons in the atomic orbitals of an atom are in their lowest , most stable energy arrangements and since Electrons must be filled following the Aufbau's principle(electrons fill lowest energy shells first)

Now, Titanium lies in period IV and  group 4 of the periodic table with 22 as its atomic number

Thus, the ground-state electron configuration of a neutral atom of titanium is 1s²2s²2p⁶3s²3p⁶4s²3d².  

Answer:

6.4×10¯³ g of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4

= 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂

Thus, 6.4×10¯³ g of O₂ is needed for the reaction.

The equilibrium constant for the reaction is K = 0.137

We obtain the equilibrium constant considering the following equilibria and their constants:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

COO(s) + CO(g) → Co(s) + CO₂(g)   K₂ = 490

We write the first reaction in the forward direction because we need H₂(g) in the reactants side:

(1)     COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):

(2)   Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

From the addition of (1) and (2), we obtain:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

+

Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

-------------------------------------------------

H₂(g) +  CO₂(g) → CO(g) + H₂O(g)

Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.

Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:

K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137

You can learn more about equilibrium constants here:

https://brainly.com/question/15118952

Answer:

V₂ = 0.154 Liters

Explanation:

Pressure => P

Volume   => V

Temperature => T

mass (moles) => n

This problem...

P₁ = 1.00 ATM     P₂ = 3.25 ATM

V₁ = 0.500L       V₂ = ?

T₁ =  constant    T₂ = T₁ = constant

n₁ =  constant     n₂= n₁ = constant

P₁V₁/n₁T₁ = P₂V₂/n₂T₂  => V₂ = V₁(P₁/P₂) = 0.500L (1.00ATM/3.25ATM) = 0.154 Liters

Question Completion with Options:

O coarse...few...rapid

O fine...few...slow

O fine...multiple...rapid

O coarse...few...slow

O fine...multiple...slow

Answer:

The choice that best completes the sentence is:

O coarse...few...slow

Explanation:

Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow.  This is because of the process that starts with  recrystallization, recovery, and nucleation before growth can occur.  While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.

Answer:

Asumiendo

"M"

es una variable

|

Usar como

un número romano

en lugar de

Suponiendo la multiplicación

|

Uso una lista en lugar de

H x×100 M n

Figura geométrica

línea

Propiedad como función

Paridad

aun

Derivado

d/dx(H x×100 M n) = 100 H M n

Integral indefinida

integral100 H M n x dx = 50 H M n x^2 + constante

Integral definida sobre una hiperesfera de radio R

integral integral integral_(H^2 + M^2 + n^2 + x^2<R^2) 100 H M n x dH dM dn dx = 0

Integral definida sobre un hipercubo de longitud de borde 2 L

integral_(-L)^L integral_(-L)^L integral_(-L)^L integral_(-L)^L 100 H M n x dx dn dM dH = 0

Explanation:

Solution :

A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.

Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.

In the context, the products including the stereochemical designations for any chirality centers starting from the  (R)-3-methylhex-1-yne as the substrate are attached below.