Give the following for SO2 and BrF5:
(a) Number of domains on central atom (b) Domain geometry (c) Molecular geometry (d) Hybridization of central atom

Answer:

For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality.

Explanation:

Answer:

When red cabbage is treated with an acid or a base, it produces anthocyanin, a water-soluble pigment that changes color. In acidic situations with a pH less than 7, the pigment turns red, whereas in alkaline (basic) situations with a pH more than 7, the pigment turns bluish-green.

Explanation:

Answer:

true

Explanation:

probably true

Answer:

B

Explanation:

So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.

Answer:

0.127M

Explanation:

Molarity of a solution = number of moles (n) ÷ volume (V)

Molar mass of Mg(NO3)2 = 24 + (14 + 16(3)}2

= 24 + {14 + 48}2

= 24 + 124

= 148g/mol

Using the formula, mole = mass/molar mass, to convert mass of Mg(NO3)2 to mole

mole = 14g ÷ 148g/mol

mole = 0.095mol

Volume = 750mL = 750/1000 = 0.75L

Molarity = 0.095mol ÷ 0.75L

Molarity = 0.127M

Answer:

a  lipped cylindrical glass container for laboratory use

Explanation:

Answer:

1. Both

2. Phosphorylation

3. Both

4. Phosphorylation

5. Oxidative.

6. Both

Explanation:

Phosphorylation only occurs in chloroplast and it involves larger electrical component. Both Phosphorylation and oxidative occurs in mitochondria and it involves proton gradient. They occur in plants to produce ATP. Oxidative involves in smaller electrical component.

Photophosphorylation is a process that captures the solar energy from the sun to transform it into chemical energy. It occurs in the chloroplast of a plant cell.

Photophosphorylation is a process of converting solar energy from the sun to ATP needed by plants and other organisms for cellular function and activity. This process takes place in the chloroplast of the plant cell and requires electrical components.

Oxidative Phosphorylation is the process of producing ATP with the help of oxygen and enzymes hence, occurs in aerobic cells. It does not need a larger electrical component.

Both phosphorylation and oxidative phosphorylation occurs in the mitochondria of plants cells and involves a proton gradient for the formation of ATP.

Therefore, oxidative phosphorylation option 5. involves a smaller electrical component, phosphorylation option 2. occurs in the chloroplast, and option 4. needs a larger electrical component.

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Answer:

The pressure will be 933.33 Kpa

Explanation:

Given that:

Volume V₁ = 200 cm³  (note, there is a mistake in the volume. It is supposed to be 200 cm³)

Pressure P₁ = 700 Kpa

Pressure P₂ = ??? (unknown)

Volume V₂ = 150 cm³

Temperature = constant

Using Boyle's law:

PV = constant

i.e.

P₁V₁ = P₂V₂

700 Kpa × 200 cm³ = P₂ × 150 cm³

P₂ = (700 Kpa × 200 cm³)/150 cm³

P₂ = 933.33 Kpa

Answer:

I has 2 double carbon carbon bonds

Answer:

7.4

Explanation:

Plasma proteins are part of the buffer systems of blood plasma. Plasma contains both positively and negatively charged amino and carboxyl groups. These compounds' charged portions can attract and link hydrogen(H) and hydroxyl ions(OH-), therefore act as buffers.

Plasma serves as a weak and ineffective buffer. The pH of a buffer should always be around 7.4 which is nearly neutral. As such we may deduce from the first experimental observation that there is no change in pH.

Answer:

35.75 days

Explanation:

From the given information:

For first-order kinetics, the rate law can be expressed as:

[tex]\mathsf{In \dfrac{C}{C_o} = -kt}[/tex]

Given that:

the rate degradation constant = 0.12 / day

current concentration C = 0.05 mg/L

initial concentration C₀ = 3.65 mg/L

[tex]\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}[/tex]

㏑(0.01369863014) = -(0.12) t

-4.29 = -(0.12)

t = -4.29/-0.12

t = 35.75 days

Answer:

Explanation:

578kj/mol

Answer:

Easy my dude let me help you out

A.In

B.27

C.73

D.49

E.56

F.56

G.114

H.180

Also with protons and electrons they equal the same atomic number

Answer:

[tex]V_2=0.894L[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the Boyle's law for an inversely proportional relationship between pressure and volume at constant temperature, as described in the problem statement:

[tex]P_2V_2=P_1V_1[/tex]

Thus, we solve for V2, final volume, to obtain the following result:

[tex]V_2=\frac{P_1V_1}{P_2} \\\\V_2=\frac{1.27atm*2.78L}{3.95atm}\\\\V_2=0.894L[/tex]

Regards!

Answer: The bond present in given compounds is as follows-

Explanation:

An ionic bond is always formed between a metal and a non-metal atom.

For example, MgF has metal magnesium and non-metal fluorine. So, an ionic bond is there in the compound MgF.

When sharing of electrons occur between atoms of different electronegativity then the bond formed is called a polar covalent bond.

For example, C-O has a polar covalent bond.

For example, Cl-Cl is a non-polar covalent bond.

Solution :

The equation is :

[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]

The number of the moles of HA os 0.00285, and the volume is 25 mL.

15 mL of the 0.0950 M NaOH is added.

The total volume of a solution is V = 25 mL + 15  mL = 40 mL

The pH of the solution is 6.50

Calculating the [tex]K_a[/tex] of HA

[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

Let s calculate the concentration of HA and NaOH

[tex]$[HA] = \frac{^nH_A}{V}$[/tex]

        [tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]

       = 0.07125 M

[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]

            [tex]$=\frac{0.001425 mol}{0.04L}$[/tex]

           = 0.0356 M

                                      [tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]

Initial conc. (M)            0.07125 M       0.0356 M            0 M

Change in conc. (M)   -0.0356 M       -0.0356 M        + 0.0356 M

Equilibrium conc. (M)   0.03565 M        0 M                0.0356 M

Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]

Now for [tex][H^+][/tex]

[tex]$[H^+] = 10^{-pH}$[/tex]

       [tex]$=10^{-6.5}$[/tex]

       [tex]$=3.16 \times 10^{-7}$[/tex]

Calculating the value of [tex]K_a[/tex],

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

     [tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]

     [tex]$=3.16\times 10^{-7}$[/tex]

Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

[tex]rate = k[P]^{2} [Q][/tex]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]

Second row:

2. Rate value:

[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]

3.Third row:

[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]

4. Fourth row:

[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]

Answer:

2NaCl + CuCO3

FeCl2 + BaSO4

CuCO3 + Ca(NO3)2

Explanation:

Presumably this is a double replacement reaction.

A+B + C+D → A+D + C+B

It seems I may be wrong so please try to work out the problem yourself to double check, keeping in mind the charges of each compound.

Explanation:

las levaduras son pequeños organismos unicelulares que se alimentan de azúcares simples y los descomponen en dióxido de carbono, alcohol (etanol, específicamente), moléculas de sabor y energía. El proceso se conoce como fermentación.

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, since the reaction taking place between sodium hydroxide and chlorine has is:

[tex]NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Which must be balanced according to the law of conservation of mass:

[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:

[tex]n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH[/tex]

Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.

Regards!

Answer: For the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression is K = [ H+] [NO2-] / [HNO2].

Explanation:

An expression that depicts the ratio of products and reactants raised to the power of their coefficients at equilibrium is called equilibrium constant.

An equilibrium constant is denoted by the symbol 'K'.

For example, the dissociation of nitrous acid in aqueous solution is as follows.

[tex]HNO_{2} \rightleftharpoons H^{+} + NO^{-}_{2}[/tex]

Hence, its expression for equilibrium constant is as follows.

[tex]K = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}[/tex]

Thus, we can conclude that for the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression is K = [ H+] [NO2-] / [HNO2].

Answer:

[tex]\%m=66.7\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.

Next, we apply the following equation to obtain the required concentration:

[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]

Regards!

Answer:

a) 0.15 mol.

b) 8.95 g.

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:

[tex]PV=nRT[/tex]

And proceed as follows:

a) Here, we solve for the moles, n,  as follows:

[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]

b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:

[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]

Regards!

Answer:

12

Explanation:

nNaCl= 4x3=12

Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

Answer:

A) [H3PO4] will increase, [KH2PO4] will decrease, and pH will slightly decrease.

Explanation:

A buffer is a solution which resists changes to its pH when a small amount of acid or base is added to it.

Buffers consist of a weak acid (HA) and its conjugate base (A–) or a weak base and its conjugate acid. Weak acids and bases do not completely dissociate in water, and instead exist in solution as an equilibrium of dissociated and undissociated species. When a small quantity of a strong acid is added to a buffer solution, the conjugate base, A-, reacts with the hydrogen ions from the added acid to form the weak acid and a salt thereby removing the extra hydrogen ions from the solution and keeping the pH of the solution fairly constant. On the other hand, if a small quantity of a strong base is added to the buffer solution, the weak acid dissociates further to release hydrogen ions which then react with the hydroxide ions of the added base to form water and the conjugate base.

For example, if a small amount of strong acid is added to a buffer solution that is 0.700 M H3PO4 and 0.700 M KH2PO4, the following reaction is obtained:

KH₂PO₄ + H+ ----> K+ + H₃PO₄

Therefore, [H₃PO₄] will increase, [KH₂PO₄] will decrease, and pH will slightly decrease.:

Answer:

1

Explanation:

1

Answer:

a

Explanation:

Answer:

Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

Explanation:

Oxidation:                            Al°(s) =>   Al⁺³(aq) + 3e⁻

Reduction:           3Ag⁺(aq) + 3e⁻ => 3Ag°(s)

_________________________________________

Net Rxn:           Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

One mole of neutral aluminum atoms (Al°(s)) undergo oxidation delivering 3 moles  of electrons to 3 moles silver ions (3Ag⁺³(aq)) that are reduced to 3 moles of neutral silver atoms (3Ag°(s)) in basic standard state 25°C; 1atm.

A balanced equation obeys the law of conservation of mass. According to the law of conservation of mass, mass can neither be created nor be destroyed. The coefficient of silver is 3.

A balanced chemical equation can be defined as the chemical equation in which the number of reactants and products on both sides of the equation are equal. The amount of reactants and products on both sides of the equation will be equal in a balanced chemical equation.

The numbers which are used to balance the chemical equation are called the coefficients. The coefficients are the numbers which are added in front of the formula.

The balanced chemical equation for the given redox reaction is given as:

Al (s) + 3 Ag⁺ (aq)  → Al³⁺ (aq) + 3Ag (s)

Thus the coefficient of silver is 3.

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