PLEASE HELP ASAP‼️‼️
A
B
C
D

Answer:

B. There is a very large percentage of C-12.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.

In such a way, the answer will be B. There is a very large percentage of C-12.

Regards!

Answer:

37.85 L

Explanation:

3.785 x 10.00 = 37.85 L

it would take 37.85 L  to fill a 10.00 tank

(sorry if im wrong pls dont report)

(hope this helps can i plz have brainlist :D hehe)

Answer:

false

Explanation:

consumers eat producers

Answer:

False. If they are talking about food today.

Answer:

pH = 3.68

Explanation:

We can solve this problem by using Henderson-Hasselbach's equation:

Assuming we have 1 L of the buffer solution then the molar concentrations of formate and formic acid would be:

We now have all required data to calculate the pH:

Answer:

See explanation and image attached

Explanation:

The reaction between benzophenone and phenylmagnesium bromide is a Grignard reaction.

A Grignard reagent is any alkyl magnesium halide compound. In this case, the Grignard reagent is phenylmagnesium bromide.

Reaction of Grignard reagent with a ketone yields all alcohol. Thus, the reaction of benzophenone with phenylmagnesium bromide yields triphenyl methanol.

The mechanism of the reaction and all electron movements are shown in the image attached to this answer.

Answer:

compound light microscope

Answer:

Density = 19.3 g/cm³

Explanation:

In order to answer this question we need to keep in mind the following definition of density:

As both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:

Answer:

2.5 × 10² ppm

Explanation:

Step 1: Given data

Step 2: Convert 0.050 g to μg

We will use the conversion factor 1 g = 10⁶ μg.

0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg

Step 3: Calculate the concentration of NaCl in ppm

The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.

5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm

Answer:250 ppm

Explanation:

Answer:

Substance 1 will diffuse at a faster rate.

Explanation:

We can solve this problem by keeping in mind Fick's law, which states:

Where:

As (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.

Thus the answer is substance 1.

Answer:

57.2 g

Explanation:

First we convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles, using its molar mass:

0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now convert 0.170 moles of Ba(ClO₄)₂ into grams, using the molar mass of Ba(ClO₄)₂:

Answer:

1) B

2) D

3) A

4) Ga

5) K

6)Po

7) Atomic size increases down the group

8) B<Al<Ga<In<Tl

9)Se<C<Ga

10) ionization energy is the energy required to remove electrons from the outermost shell of an atom.

Explanation:

In the periodic table, the properties of elements reoccur ''periodically'' throughout the table, hence the name 'periodic table'.

Ionization energy increases across the period hence the noble gas He has the highest ionization energy.

Since ionization energy increases across the period, group 1 elements possess the lowest ionization energy.

Since atomic size increases down the group and decreases across the period, gallium is smaller than indium, potassium is smaller than caesium, polonium is smaller than titanium and iodine is larger than bromine.

This explanation above justifies the order of increasing atomic radius of group 13 elements shown in answer number 8 above.

Since atomic size decreases across the period, the order of increasing atomic size shown in answer number 9 above is correct.

Ionization energy is the energy required to remove electrons from the outermost shell of an atom.

Answer:

C₁₅N₃H₅

Explanation:

Let's assume we have 240 g of the dye (1 mol), in that case we'd have:

Now we convert the masses of each element into moles, using their respective molar masses:

Thus the molecular formula is C₁₅N₃H₅.

Answer:

The initial temperature of the water is -75.08 K.

Explanation:

Given that,

Mass of water, m = 27 g

Heat absorbed, Q = 1,5000 J

Final temperature of water, T₂ = 57.7 K

The specific heat of water is 4.184 J/​g-K

We know that,

[tex]Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\T_1=T_2-\dfrac{Q}{mc}[/tex]

Put all the values,

[tex]T_1=57.7-\dfrac{15000}{27\times 4.184 }\\\\=-75.08\ K[/tex]

So, the initial temperature of the water is -75.08 K.

Answer:

Explanation:

As the measurements are given to us in centimeters, let's start by calculating the surface area in square centimeters:

Now we convert 100.75 cm² to m², as follows:

Finally we convert 0.010075 m² to nm², as follows:

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of [tex]-6.4^oC[/tex] and a molal freezing point depression constant [tex]K_f= 3.96^oC.kg/mol[/tex]. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at [tex]-13.6^oC[/tex]. Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: 129.66 g of glycine will be dissolved.

Explanation:

Depression in the freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m[/tex]

                                 OR

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times \frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (g)}}[/tex]        ....(1)

where,

i = Van't Hoff factor = 1 (for non-electrolytes)

Freezing point of pure solvent = [tex]-6.4^oC[/tex]

Freezing point of solution = [tex]-13.6^oC[/tex]

[tex]K_f[/tex] = freezing point depression constant  = [tex]3.96^oC/m[/tex]

[tex]M_{solute}[/tex] = Molar mass of solute (glycine) = 75.07 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 950 g

Plugging values in equation 1:

[tex]-6.4-(-13.6)=1\times 3.96\times \frac{\text{Given mass of glycine}\times 1000}{75.07\times 950}\\\\\text{Given mass of glycine}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\\text{Given mass of glycine}=129.66g[/tex]

Hence, 129.66 g of glycine will be dissolved.

Answer:

A positively charged nucleus consisting of protons and neutrons

Explanation:

The atomic nucleus is a positively charged region located at the core of an atom that consists of positively charged protons and neutral neutrons while the negatively charged electrons make up the outer cloud (electrons are therefore not contained in the nucleus).

Explanation:

[tex]q = 350 \times0.897 \times (70 - 25) \\ q = 14127.75[/tex]

Answer:

[Ar] 4s²3d³

Explanation:

Vanadium has atomic number of 23. The electronic configuration of vanadium can be written as:

V (23) => 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d³

NOTE: After the 18th electron, 4s will be filled before 3d.

We can also write the electronic configuration of an element in its condensed form by writing the symbol of the noble before the desired element in a squared bracket followed by the remaining electrons to complete the electronic configuration of the element.

The electronic configuration of vanadium in its condensed form is given below:

The noble gas before vanadium is Argon (Ar) with atomic number of 18. Thus, the electronic configuration of vanadium becomes:

V (23) => [Ar] 4s²3d³

Answer:

(d) the ion with 24 electrons, 30 neutrons, and a 3+ charge ... (b) the ion with 36 electrons, 35 protons, and 45 neutrons.

Answer: 6

Explanation:

You would need 6

Answer:

Explanation:

The missing image can be seen below.

From the given information:

The elimination process follows E2 mechanism which is a 2nd order kinetics.

At E2 mechanism, the base attaches with the beta hydrogen while also removing the leaving group in the same process. In the given compound 2-chloro-2-methylpropane, chloride is the leaving group that results in the product; 2-methylprop-1-ene.

The mechanism is seen in the second image,

Answer:

verdadero

Explanation:

porque esoo [tex]\lim_{n \to \infty} a_n x_{123} \frac{x}{y} \sqrt[n]{x} x^{2} \sqrt{x} \pi \neq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right.[/tex] 

Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

[tex]Heat of combustion = heat capacity of calorimeter * deltaT\\[/tex]

Given,

The heat of combustion = 10.5kJ/g.

[tex]deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC[/tex]

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

[tex]deltaT =heat capacity of calorimeter * (change in temperature)\\10.5kJ/g = heat capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC[/tex]

Answer:

The heat capacity of the calorimeter is [tex]3.0kJ/g.^oC.[/tex]

Answer:

a) 320: two significant figures.

b) 2,366: four significant figures.

c) 73.0: three significant figures.

d. 532.5: four significant figures.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to write each number by knowing we move the decimal places to the right as much as the exponent is, and also, we count every figure, even zeros, because they are to the right of the first nonzero digit:

a) 320: two significant figures because the rightmost zero is not preceded o followed by a decimal place.

b) 2,366: four significant figures.

c) 73.0: three significant figures, because the zero is followed by the decimal place.

d. 532.5: four significant figures.

Regards!

Explanation:

Aromatic nitrogen bonded to hydrogen, as found in pyrrole must be represented as [nH] ; thus imidazole is written in SMILES notation as n1c[nH]cc1 . When aromatic atoms are singly bonded to each other, such as in biphenyl, a single bond must be shown explicitly: c1ccccc1-c2ccccc2 .

Answer:

0.01684 s⁻¹

Explanation:

In a first-order reaction, the rate is proportional to the concentration of only one reactant (raised by 1). In this case, the reactant is H₂O₂. Thus, the rate law is the following:

rate = k [H₂O₂]

We have the following data for the initial rate:

rate = 0.00842 mol/(L·s)

[H₂O₂] = 0.500 mol/L

So, we introduce the data in the expression for the rate law to calculate k:

k = rate/[H₂O₂] = (0.00842 mol/L·s)/0.500 mol/L = 0.01684 s⁻¹