net ionic equation of HF(aq)+RbOH(aq)=H2O(l)+RbF(aq)

Answer:the first column has one valance electron the second has 2 and so on.

The only exception is helium that has 2.

Explanation:

Answer:

Mass =   51 g

Explanation:

Given data:

Mass of ammonia formed = ?

Number of moles of hydrogen = 4.50 mol

Solution:

Chemical equation:

N₂ + 3H₂     →     2NH₃

Now we will compare the moles of hydrogen and ammonia.

                H₂           :            NH₃

                  3           :              2

                 4.50       :           2/3×4.50 = 3 mol

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 3 mol × 17 g/mol

Mass =   51 g

Answer:

450 grams of Na2SO4

Explanation:

hope that helps

Answer:

electronic configuration: 1s^2,2s^2,2p^3            

name: nitrogen

unpaired electron: 3

Explanation:

The atom with 2 electrons and 5 electrons in the first and second energy level is NITROGEN. The element has 3 unpaired electron.

Therefore, the 2p orbital has 3 unpaired electrons in it's outermost shell.

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Answer:

tbh idekkk

Explanation:

Answer:

D. 3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

Explanation:

The complete ionic equation includes all the ions and the insoluble species. Let's consider the following complete ionic equation.

3 Co²⁺(aq) + 6 NO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble species. The corresponding net ionic equation is:

3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

Answer:  Its "C"

Explanation:

Appearance and density

sorry this took 14 hours lol

pls mark brainiest

Answer:

fluorine atom

Explanation:

For the extraction of at least 95% of Y in water, 3 extractions are to be performed.

Distribution coefficient can be defined as the ratio of the concentration of solute in an organic solvent to water.

Distribution coefficient = [tex]\rm \dfrac{concentration\;in\;solvent}{concentration\;in\;water}[/tex]

Concentration in water = 100-X per 50 ml

Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]

4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]

4 = [tex]\rm \dfrac{50X}{1000-X}[/tex]

4000 - 4X = 50X

X = 74.1 %

Thus, after the first extraction, the amount of Y extracted is 74.1%.

We have to extract at least 95% of Y. Thus, the second extraction is performed.

Remaining y = 25.9%

The concentration in organic solvent = X per 10 ml

Concentration in water = 25.9 -X

Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]

4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]

4 = [tex]\rm \dfrac{50X}{259-X}[/tex]

1036 - 4X = 50X

X = 19.2%

Thus, after the second extraction the amount of Y extarcted = first extraction + second extraction

The amount of Y extracted = 74.1 + 19.2 %

The amount of Y extracted = 93.3%

The remaining Y for third extraction = 100 - 93.3

The remaining Y for the third extraction = 6.7%

Concentration in water = 100 - 6.7

Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]

4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]

4 = [tex]\rm \dfrac{50X}{67-X}[/tex]

268 - 4X = 50X

X = 5.0%

The total extraction after third extraction = 93.3 + 5%

The total extraction after third extraction = 98.3%.

Thus for the extraction of at least 95% of Y in water, 3 extractions are to be performed.

For more information about the distribution coefficient, refer to the link:

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Answer:

pH = 4.89

0.6171g of KOH must be added

Explanation:

Sodium acetate reacts with HCl producing acetic acid as follows:

NaC2H3O2 + HCl → HC2H3O2 + NaCl

That means the moles added of HCl are moles of acetic acid produced and moles of acetate are initial moles - moles of HCl

Moles HCl = Moles acetic acid:

0.204L * (0.452mol / L) = 0.0822 moles acetic acid.

Initial moles sodium acetate:

0.500L * (0.400mol / L) = 0.200 moles

Moles sodium acetate:

0.200 moles - 0.0822 moles = 0.1172 moles sodium acetate

The pH of this buffer (Mixture of a weak acid and its conjugate base) is obtained using H-H equation:

pH = pKa + log [A-] / [HA]

Where pH is pH of the buffer,

pKa is pKa of the buffer (4.74 for acetic acid)

[A-] Moles of sodium acetate -Conjugate base-

And HA moles of acetic acid -Weak acid-

pH = 4.74 + log [0.1172 moles] [0.0822 moles]

In 0.500L the moles of the buffer are:

0.500L * (0.200moles / 0.704L) = 0.142 moles of buffer

For a pH of 0.15 units more (4.89 + 0.15 = 5.04):

5.04 = 4.74 + log [Acetate] / [Acetic acid]

0.3 = log [Acetate] / [Acetic acid]

1.9953 = [Acetate] / [Acetic acid] (1)

And as:

0.142 = [Acetate] + [Acetic acid] (2)

Replacing (2) in (1):

1.9953 = 0.142 - [Acetic acid]  / [Acetic acid]

1.9953 [Acetic acid] = 0.142 - [Acetic acid]

2.9953 [Acetic acid] = 0.142 moles

[Acetic acid] = 0.0474 moles of acetic acid must remain after the addition of KOH.

In the beginning, moles of acetic acid are:

0.0822 moles * (0.500L / 0.704L) = 0.0584 moles.

That means moles added of KOH (Equal to moles of acetic acid that react) are:

0.0584 moles - 0.0474 moles = 0.0110 moles of KOH

In grams (Molar mass KOH = 56.1g/mol):

0.0110 moles KOH * (56.1g/mol) =

Answer:

Ionic compounds generally form between elements that are metals and elements that are nonmetals. For example, the metal calcium (Ca) and the nonmetal chlorine (Cl) form the ionic compound calcium chloride (CaCl2). In this compound, there are two negative chloride ions for each positive calcium ion.

Answer:

6.36  g AgCl

Hope this helps!

Answer:

el n = M/m. El nº de moles = m/M

Se utiliza la fórmula general de los gases: P*V = n*R*T

P = 760/760 = 1 atm

V = 0,5 L

n = masa de Cl2/masa molecular = masa de Cl2/70

R = 0,082

T = 273 + 20º C = 293 K

1*0,5 = masa de Cl2/70*0,082*293

0,5/0,082*293 = masa de Cl2/70

0,0208 = masa de Cl2/70

masa de Cl2 = 70*0,0208 = 1,46 g

epilacion:

Suerte

Answer:

MgS Magnesium sulfide

Answer:

Explanation:

Answer:

b. lead

Explanation:

look on the periodic table and it has its atomic mass and lead (Pb) is the heaviest

Answer:

b) 4

Explanation:

The electron transport chain may be defined as the series of the complexes which transfer the electrons from electron donors to the electron acceptors through a redox reactions, and also couples the electron transfer along with the transfer of the protons across the membrane.

The electron transport chain or the ETC is made up of peptides and enzymes, and some other molecules.

The full form of ATP is Adenosine triphosphate. It is a chemical compound that provides energy to the living cells.

From one run in the ETC or the electron transport chain, 4 ATP can be generated.

Answer:

(a). Kindly check the explanation section.

(b). 24.5 M^-1 cm^-

(c). 0.0318 M

Explanation:

So, we are given the following data parameters which is going to aid us in solving the above Question;

The path length = 1cm, extinction coefficient = unknown, absorbance is unknown and concentration= 0.02mM.

Using the formula below; we can determine the extinction coefficient as;

Absorbance = extinction coefficient × concentration × path length. -----------(1).

(a). Since, the absorbance is not given in the Question it won't be possible to determine the value of the extinction coefficient. Thus, say the value of absorbance = A.

Then, extinction coefficient = A/ 0.02 × 1 cm.

(b). Making use of the formula above, the extinction coefficient can be Determine as follows;

extinction coefficient = 1.84/0.075 × 1 = 24.5 M^-1 cm^-1.

(c). The concentration can be Determine by also making use of the formula above and making the concentration the subject of the formula;

Concentration = absorbance/ extinction coefficient × path length.

Concentration = 0.78/24.5 M^-1 cm^-1. × 1cm = 0.0318 M.

Answer:

they use it when a living organism dies.

Explanation:

when d organism decomposes carbon is released that enters back into d earth. this follows d law of mass

Answer: C

Explanation:

A. As they do not have a given shape and expand to fill the container, modifying the size of the container modifies the space the gas occupies

B. They are nearly free molecules, which means no bound between them, moving fast and freely around the container

C. NOT A PROPERTY. As said before, as there are no bindings between molecules and move freely, they cannot have a fixed size cause they're always on the move.

D. Because they expand to fill the container, it's easy to them to occupy more space than in any other state, as the container is the only thing that defines how much volume they occupy

Answer:

Bromine is a very reactive element. While it is less reactive than fluorine or chlorine, it is more reactive than iodine. It reacts with many metals, sometimes very vigorously. For instance, with potassium, it reacts explosively.

Explanation:

Answer: Bromine is a high reactive element. While it is not as reactive as fluorine or chlorine, it is still more reactive than iodine. It reacts with many metals, and even sometimes very aggressively. For example, when mixed with potassium, it will explode.  

Explanation:

Answer: increase number of collisions

Explanation:

Decreasing the volume increases the likelihood of particles interacting more often.

Answer:

Atoms collided more often when the volume was lower.

Explanation:

plato

Answer:

yes

Explanation:

Answer:

yes

Explanation:

let's say you have ice, you put it on a pan then it heats up you get a liquid after a while it turns into a gas

Answer:

b.  growth in the size of the mineral grains

Explanation:

Non-foliated texture shown by a metamorphic change is depicted by growth in the size of the mineral grains.

Answer:

its b home slice

Explanation:

Answer:

See Explanation

Explanation:

Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.

This implies that this compound Pb2O3  (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.

Hence Pb2O3  contains PbO and PbO2 units.

Answer:

Answer: Tin (II) nitride

Answer: Some salt is trapped between ice crystals, but most are left behind in the unfrozen seawater.

Explanation:

Answer: The salinity of the water increases

Explanation: Hope this helps!

Answer:

Explanation:

In the equivalent mass measurement , one gram equivalent of metal produces 11200 mL of hydrogen gas .

11200 mL hydrogen gas is produced by 1 gram equivalent of gas at STP

11200 mL hydrogen gas is produced by 20 g  of gas at STP

80 mL hydrogen gas is produced by 20 x 80 / 11200  g  of gas at STP

= 142.85 mg .

= 140 mg . ( rounded to 2 sig figures )

Answer:

Explanation:

pH = 4.55

[ H⁺ ] = 10⁻⁴°⁵⁵

= 2.82 x 10⁻⁵

Let the acid be HA and its molar concentration be a .

         HA         ⇄       H ⁺       +        A ⁻

a - .094x10⁻³a      .094x10⁻³a    .094x10⁻³a

.094x10⁻³a  = 2.82 x 10⁻⁵

a = 30 x 10⁻²

= .3

a - .094x10⁻³a   = .3 - .094 x 10⁻³ x .3

= .29997 approx

Ka =  2.82 x 10⁻⁵ x  2.82 x 10⁻⁵ /  .29997

= 7.95 x10⁻¹⁰ / .29997

= 26.5 x 10⁻¹⁰

= 27 x 10⁻¹⁰ ( rounding off to two digits )

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}\ \text{mol}^{-1}[/tex]

For the 4.12 g sample

Moles of a substance is given by

[tex]n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}[/tex]

Number of molecules is given by

[tex]nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}[/tex]

For the 19.37 g sample

[tex]n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}[/tex]

Number of molecules is given by

[tex]nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}[/tex]

[tex]1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}[/tex]

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.