At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.

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**

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:

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:

kjajjahahayq :/

Explanation:

a sbywsbgv usnwbhx hg xw nx hb gs

Answer:

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.  

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:

pH = 10.20

Explanation:

The HClO reacts with NaOH as follows:

HClO + NaOH → H2O + NaClO

Where HClO and NaOH react in a 1:1 reaction.

As the concentration of both reactions is the same and the reaction is 1:1, to reach equivalence point are required the same 25.0mL.

And the NaClO produced decreases its concentration in 2 because the volume is doubled.

The concentration of NaClO is: 0.150M / 2 = 0.075M

The equilibrium of NaClO is:

NaClO(aq) + H2O(l) ⇄ HClO(aq) + OH-(aq)

Where Kb of reaction is 1.0x10⁻¹⁴ / Ka =

1.0x10⁻¹⁴ / 3.0x10⁻⁸ = 3.33x10⁻⁷ = [HClO] [OH-] / [NaClO]

[NaClO] = 0.075M

As both HClO and OH- comes from the same equilibrium,

[HClO] = [OH-] = X

Where X is the reactoin coordinate

Replacing:

3.33x10⁻⁷ = [X] [X] / [0.075M]

2.5x10⁻⁸ = X²

X = 1.58x10⁻⁴M = [OH-]

pOH = -log [OH-]

pOH = 3.80

pH = 14 - pOH

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!

3,3-dimethylhexane is the nomenclature of the compound

Answer:

a. 1.12 L

Explanation:

Step 1: Write the balanced equation for the photosynthesis

6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)

Step 2: Calculate the moles corresponding to 2.20 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

2.20 g × 1 mol/44.01 g = 0.0500 mol

Step 3: Calculate the moles of O₂ produced

The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol

Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP

At STP, 1 mole of O₂ occupies 22.4 L.

0.0500 mol × 22.4 L/1 mol = 1.12 L

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:

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:

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:

Mg> H> Cu

Explanation:

We can see from the question that hydrochloric acid reacted with magnesium as follows;

Mg(s) + 2HCl(aq) ----> MgCl2(aq) + H2(g)

Copper does not react with HCl which means that copper is less reactive than hydrogen hence it can not displace hydrogen from a dilute acid solution.

The order of reactivity of the elements then is ; Mg> H> Cu

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

Explanation:

here's the answer to your question

Answer:

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

Explanation:

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:

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:

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:

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:

hydrogen

Explanation:

The gas with the least molecular weight effuses the fastest (Graham's Law). Hence, H gas has a higher rate of diffusion compared to N, O, and Cl.

So, Cl is the slowest when it comes to the rate of diffusion, because it has the highest molecular weight.

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:

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.

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:

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.  

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

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: