True or False
Helium belongs to Noble Metals

Answers

Answer 1

Answer:

This answer is "True"

Answer 2
the answer would be *true*

Related Questions

This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and the partial pressures of the gases

Answers

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

Where [] are concentrations at equilibrium

And Q is:

Q = [HI]² / [H2] [I2]

Where [] are actual concentrations of the reactants.

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

Which compound is insoluble in water?

Answers

Answer:

The answer is C... I am almost positive.

If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give your reason.​

Answers

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!

For the reaction...
N2 + O2 <=> 2NO: AH = +182 kJ mol-1.
If the temperature is increased the equilibrium position will shift
Your answer:
a) to the left
b) to the right
c) to the left and right
d) neither left nor
right

Answers

Answer:

B

Explanation:

AH is positive so the forward reaction is endothermic. Thus, increasing temperature would cause equilibrium to shift to the right as endothermic reaction favors higher temperature. This increases the yield of NO.

Answer: B

Explanation:

You dissolve 14 g of Mg(NO3)2 in water and dilute to
750 mL. What is the molarity of this solution?

Answers

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

What are the laws and calculations governing gas behavior?

Answers

Answer:

Laws governing gas behavior.

Explanation:

Boyle's law:

It relates the pressure and volume of an ideal gas at a constant temperature.

According to this law:

"The volume of a fixed amount of gas at constant temperature is inversely proportional to its pressure".

[tex]P \alpha V[/tex].

Charle's law:

It relates the volume and absolute temperature of an ideal gas at a constant pressure.

According to this law:

"The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature".

[tex]V \alpha T[/tex].

Avogadro's law:

According to this law:

equal volumes of all gases under the same conditions of temperature and pressure contain, an equal number of moles.

[tex]V \alpha n[/tex].

Ideal gas equation:

By combining all the above-stated gas laws, this equation is formed as shown below:

[tex]V \alpha \frac{nT}{P} \\=> V= R. nT/ P\\=>PV=nRT[/tex]

R is called universal gas constant.

It has a value of 0.0821L.atm.mol-1.K-1.

Answer:

Boyle's law, Charle's law,  Guy Lussac's law and Avogadro's law

Explanation:

All the gases behaves similarly when the environment conditions are normal. But when the physical condition changes like when the pressure, volume or temperature changes, the gas behaves differently and shows a deviation.

The number of gas laws are :

Boyle's Law

Boyle's law states that when the temperature remaining constant, the pressure of the gas varies inversely to the volume of the gas.

i.e.   [tex]P \propto \frac{1}{V}[/tex]

Charle' law

Charle's law states that when pressure is constant, the temperature of a gas is directly proportional to the volume.

i.e. , [tex]$T \propto V$[/tex]

Gay Lussac's law

Gay - Lussa law states the volume and the mass of the pressure of the gas is directly proportional to the  temperature of the gas.

i.e. P.T = constant

Avogadro's law

It states that under the conditions of same pressure as well as temperatures, the gases having equal volumes will have same numbers of molecules.

i.e. [tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex] = constant

Please help me, it’s my last try

Answers

Answer:

Group 1A: alkali metals, or lithium family.

Group 2A: alkaline earth metals, or beryllium family.

Group 7A: the manganese family.

Group 8A: the iron family.

Explanation:

Answer:

1A: Alkali Metals

2A: Alkaline Earth Metals

7A: Halogens

8A: Noble Gases

4. Complete the following equations:
CuCl2 + Na2CO3 → 2 NaCl +............
FeSO4 + BaCl2 →


Cu(NO3)2 + CaCO3

Answers

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.

La función de la levadura en quimica

Answers

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.

A pressure cooker contains 5.68 L of air at a temperature of 390 4K if the absolute pressure of the air in the pressure cooker is 205 Pa how many moles of air are in the cooker

Answers

Answer:

3.59x10⁻⁴ mol

Explanation:

Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:

P = 205 PaV = 5.68 Ln = ?R = 8314.46 Pa·L·mol⁻¹·K⁻¹T = 390.4 K

We input the data given by the problem:

205  Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 K

And solve for n:

n = 3.59x10⁻⁴ mol

The compound sodium hydrogen sulfate is a strong electrolyte. Write the reaction when solid sodium hydrogen sulfate is put into water:

Answers

Answer:

NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)

Explanation:

Sodium hydrogen sulfate is a strong electrolyte, that is, when dissolved in water it completely dissociates into the cation sodium and the anion hydrogen sulfate. The corresponding chemical equation is:

NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)

4) 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.
rate = k[P]?[Q]
Complete the table of data below for the reaction between P and Q

*Help asap please*

Answers

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]

The information below describes a redox reaction.
Ag+ (aq) + Al(s) — Ag(s) + Al3+ (aq)
Ag+ (aq) + -> Ag(s)
Al(s)->A3+ (aq) + 3e-
What is the coefficient of silver in the final, balanced equation for this reaction?

Answers

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.

What is a balanced equation?

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.

To know more about balanced equation, visit;

https://brainly.com/question/29769009

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which type of chemical bond would be formed between two elements having electron configuration of 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4​

Answers

The electron configuration
1
s
2
2
s
2
2
p
6
3
s
2
3
p
2
is the element Silicon.
The key to deciphering this is to look at the last bit of information of the electron configuration
3
p
2
.
The '3' informs us that the element is in the 3rd Energy Level or row of the periodic table. The 'p' tells us that the element is found in the p-block which are all of the Groups to the right of the transition metals, columns 13-18. The superscript '2' tells us that the element is found in the 2nd column of the p-block Group 14.

In the presence of excess iodide ions, the iodine formed by reaction of iodide with NBS will react further to form triiodide ions. What does the triiodide combine with to form the blue color of the endpoint

Answers

Answer:

Starch.

Explanation:

When the triiodide combine with starch, it forms dark blue colour. Amylose in starch is responsible for the occurrence of a deep blue color when the iodine is combine with the starch. The iodine molecule goes inside of the amylose coil which makes a linear triiodide ion complex that goes into the coil of the starch that leads to an intense blue-black color in the end so we can say that starch turns the colour into blue.

1) 7.269 moles of oxygen gas are used in combusting butane (C H..). How many moles of carbon dioxide
gas are produced? You must start with a balanced chemical equation. Start with a balanced equation

Answers

Explanation:

C4H10 + 13/2O2 ---------> 4CO2 + 5H2O

so u can work out the amount of moles by doing

moles=mass/mr

mr of C4H10 is 12 × 4 + 10 =58

=7.269/58

= 0.125moles

Then u can use the molar ratio which is

6.5:4

0.125 ÷6.5 × 4 = 0.0769moles

hope this helps:)

a polluted lake is 0.300 μg (micrograms) per liter of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 15.0 square miles and an average depth of 27.0 feet?

Answers

Answer:

95.9 kg

Explanation:

First we convert 15.0 mi² to m²:

15.0 mi² * ([tex]\frac{1609.34 m}{1mi}[/tex])² = 3.88x10⁷ m²

Then we convert 27.0 ft to m:

27.0 ft * [tex]\frac{0.3048m}{1ft}[/tex] = 8.23 m

Now we calculate the total volume of the lake:

3.88x10⁷ m² * 8.23 m = 3.20x10⁸ m³

Converting 3.20x10⁸ m³ to L:

3.20x10⁸ m³ * [tex]\frac{1000L}{1m^3}[/tex] = 3.20x10¹¹ L

Now we calculate the total mass of mercury in the lake, using the given concentration:

0.300 μg / L * 3.20x10¹¹ L = 9.59x10¹⁰ μg

Finally we convert μg to kg:

9.59x10¹⁰ μg * [tex]\frac{1kg}{1x10^9ug}[/tex] = 95.9 kg

Phosphine, PH3, a reactive and poisonous compound, reacts with oxygen as follows: 4PH3(g) 8O2(g) - P4O10(s) 6H2O(g) If you need to make 6.5 moles of P4O10, how many moles of PH3 is required for the reaction

Answers

Answer: 26 moles of [tex]PH_3[/tex] are required for the reaction.

Explanation:

We are given:

Moles of [tex]P_4O_{10}[/tex] = 6.5 moles

The given chemical reaction follows:

[tex]4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)[/tex]

By the stoichiometry of the reaction:

If 1 mole of [tex]P_4O_{10}[/tex] is produced by 4 moles of [tex]PH_3[/tex]

So, 6.5 moles of [tex]P_4O_{10}[/tex] will be produced by = [tex]\frac{4}{1}\times 6.5=26mol[/tex] of [tex]PH_3[/tex]

Hence, 26 moles of [tex]PH_3[/tex] are required for the reaction.

which of the following illustrates a reversible change a cooking corn be rusting c frying egg and the boiling water​

Answers

The answer is : Boiling water

Boiling water is a reversible change because you can take the water off the heat, and it will return to room temperature or it’s regular temperature.

20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.

Answers

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!

Rocks are classified as igneous, metamorphic, or sedimentary according to

Answers

Answer:

D. the minerals they contain

Hope this answer is right!!

a) If we have a 4.5 L container of CH 10 gas at a temperature of 178 K and a pressure of 0.50 atm, then how many moles of CaHio do
we have?
b) How many grams of C4H1o do we have?

Answers

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!

At a fixed volume, a four-fold increase in the temperature of a gas will lead to _______ in pressure.
Question 2 options:

A)

no change

B)

a two-fold decrease

C)

a four-fold decrease

D)

a four-fold increase

Answers

Answer:

D) a four-fold increase

Explanation:

According to Gay-Lussac's law, which states that the pressure of a given amount of gas is directly proportional to the temperature at a constant volume, the pressure increases with an increase in temperature.

According to this question, at a fixed volume, a four-fold increase in the temperature of a gas will lead to a four-fold increase in the pressure as well.

38. Consider the following equilibrium:
2CO(g) + O2(g) =2CO2
Keg=4.0 x 10-10
What is the value of Key for 2CO2(g) + 2COR + O2g) ?​

Answers

Answer:

[tex]Key=2.5x10^{-9}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the equilibrium constant value for the reverse reaction:

[tex]2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g)[/tex]

By knowing that the equilibrium expression is actually:

[tex]Key =\frac{[CO]^2[O_2]}{[CO_2]^2} =\frac{1}{Keg}[/tex]

Thus, we plug in and solve for the inverse of Keq to obtain Key as follows:

[tex]Key =\frac{1}{4.0x10^{-10}}\\\\Key=2.5x10^{-9}[/tex]

Regards!

calculate the maximum theoretical percent recovery from the recrystallization of 1.00g of benzoic acid

Answers

Answer:

The maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%

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

The solubility of benzoic acid in water is 6.80g per 100mL at 100 degrees C and 0.34 g per 100mL at 25 degrees C.

Calculate the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water, assuming the solution is filtered at 25 degrees C.

Explanation:

Solubility of benzoic acid in water at 100 degrees C = 6.80g per 100mL

Solubility of benzoic acid in water  at 25 degrees C =  0.34 g per 100mL

Mass of benzoic acid to be theoretically recovered from 100 mL of water = 6.80 g - 0.34 g = 6.46 g

At 25 degrees;

0.34 g of benzoic acid is present in 100 mL of water

x g of  benzoic acid will be present in 15 mL of water

x = 0.34 × 15 / 100 = 0.051 g

Mass of benzoic acid to be theoretically recovered from 25 mL of water = 1.00 g - 0.051 g = 0.949 g

Maximum theoretical percent recovery = (mass recovered / original mass dissolved) x 100%

Maximum theoretical percent recovery =  (0.949 / 1.00) × 100% = 94.9 %

Therefore, the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%

In the reaction below, what is the limiting reactant when 1.24 moles NH3 of reacts with 1.79 moles of NO?

4NH_3 + 6NO (right arrow) 5N_2 + 6H_2O

1. NO
2. H_2O
3. NH_3
4. N_2

Answers

Answer:

Option 1. NO

Explanation:

The balanced equation for the reaction is given below below:

4NH₃ + 6NO —> 5N₂ + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Therefore, 1.24 moles of NH₃ will react with = (1.24 × 6)/4 = 1.86 moles of NO

From the calculation made above, we can see that a higher amount of NO (i.e 1.86 moles) than what was given (i.e 1.79 moles) is needed to react completely with 1.24 moles of NH₃.

Therefore, NO is the limiting reactant and NH₃ is the excess reactant.

Thus, the 1st option gives the correct answer to the question

Answer:

1. NO .

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to identify the limiting reactant by simply calculating the moles of any product, say N2, via the moles of each reactant and including the corresponding mole ratio (4:5 and 6:5):

[tex]1.24molNH_3*\frac{5molN_2}{4molNH_3}=1.55molN_2 \\\\1.79molNO*\frac{5molN_2}{6molNO}=1.50molN_2[/tex]

Thus, since NO yields the fewest moles of N2 product, we infer it is the limiting reactant.

Regards!

Define pure substance. How is it classified on the basid of chemical properties?​

Answers

Answer:

if it is pure, the substances is either an element or a compound. if a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. if its composition is uniform throughout, it is a homogeneous.

Acetylide ions react with aldehydes and ketones to give alcohol addition products.

a. True
b. False

Answers

Answer:

a

Explanation:

Atoms are found to move from one lattice position to another at the rate of 300,000 jumps/s at 500 0C when the activation energy for their movement is 10,000 cal/mol. Calculate the jump rate at 400 0C.

Answers

Answer:

1

Explanation:

1

If 4.00 moles of O2 occupies a volume of 5.0 L at a particular temperature and pressure, what volume will 3.00 moles of oxygen gas occupy under the same condition?

Answers

Answer: Volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.

Explanation:

Given: [tex]n_{1}[/tex] = 4.00 moles,          [tex]V_{1}[/tex] = 5.0 L

[tex]n_{2}[/tex] = 3.00 moles,                 [tex]V_{2}[/tex] = ?

Formula used is as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{5.0 L}{4.00 mol} = \frac{V_{2}}{3.00 mol}\\V_{2} = 3.75 L[/tex]

Thus, we can conclude that volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.

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