The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?

Answer:

I think you can make 26, hope this helped.

Explanation:

Explanation:

example is copper iron...........

D.22

is my answer than welcome

Answer:

Please find the complete solution in the attached file.

Explanation:

Answer:

1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

Explanation:

Let's consider the dissolving equations for the following compounds.

1. Rubidium hydroxide

RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Sodium carbonate

Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. Ammonium selenite

(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

Answer:

Its atomic number is 14 and its atomic mass is 28. The most common isotope of uranium has 92 protons and 146 neutrons. Its atomic number is 92 and its atomic mass is 238 (92 + 146).

Answer:

The correct answer is C. Nucleic acid

Explanation:

Nucleic acids are biological polymers which play an important role in the storage and expresion of genetic information. There are two types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are basically composed of:

- a 5-carbon sugar: deoxyribose in DNA and ribose in RNA

- phosphate group

- a nitrogenous base: adenine, cytosine, guanine and thymine in DNA; while RNA contains adenine, cytosine, guanine and uracil.

Answer:

a) Hence, T = 207 K.

b) Hence, T2 = 226 K.

Explanation:

Now the given,

n = 0.27 moles ; P = 2.5 atm ; T = 298 K

a) γ = 5/3 since Ne is a monoatomic gas.

[tex](1 - \gamma )/\gamma = -2/5\\T1 P1^{(1-\gamma)/\gamma}=T2 P2^{(1-\gamma)/\gamma}\\T2 = T1(P1/P2)^{(1 - \gamma)/\gamma}\\T2 = 298 (2.5/1)^{-2/5}= 207 K\\[/tex]

Hence, T = 207 K

b) We know that,[tex]U = W = n Cv (T2 - T1) = -P (V2 - V1)[/tex]

[tex]n(3/2)R(T2 - T1) = -P( n R T2/P2 - n R T1/P1)\\3/2(T2 - T1) = -P (T2/P2 - T1/P1)[/tex]

But P = P2

[tex]3/2(T2 - T1) = -P2(T2/P2 - T1/P1)\\3/2(T2 - T1) = -T2 + P2T1/P1[/tex]

This gives us:

[tex]T2 = 2/5(P2/P1 + 3/2)T1\\T2 = 2/5 x (1 /2.5 + 3/2)/(298)\\T2 = 19/25 x 298 = 226 K[/tex]

Hence, T2 = 226 K

Answer:

Because sucrose is a complex disaccharide, it is not classified as either an aldose or a ketone. Instead, it is a compound that contains both. glucose is aldose sugar and fructose is a ketose sugar.

Answer:

sosksjsjjs

Explanation:

even i know how to type şïllily

Answer:

A. 3-chloro-1-methylcyclobutane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the name of this compound is A. 3-chloro-1-methylcyclobutane because of the fact that the parent chain is a cyclobutane which starts by the methyl radical as it has the priority over the chlorine radical which is actually named first at the third carbon (clockwise).

Therefore the name is given in A, accordingly to the IUPAC rules of nomenclature.

Regards!

Explanation:

the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet

Answer:

please translate in english

In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.

Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase

Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.

The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids

Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

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Answer:

R.F.M of Iron (II) oxide :

[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]

Moles :

[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]

Molecules :

[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]

The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.

Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.

The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.

Given, the mass of the iron oxide = 35.2 ×10⁻²³ g

The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol

71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³

35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)

The number of molecules of FeO in a given mass = 2.95 molecules

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Answer:

[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:

[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]

Now, we can separate the nitrates in ions as they are aqueous to obtain:

[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]

And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:

[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]

Best regards!

Answer:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Double replacement reaction.

It is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Explanation:

Hello there!

In this case, according to the given information, it turns possible for us to solve this problem by firstly considering that this reaction occurs between potassium iodide and lead (II) nitrate to yield potassium nitrate and lead (II) iodide which is clearly not balanced since we have one iodine atom on the reactants and two on the products, that is why the balance implies the placement of a coefficient of 2 in front of both KI and KNO3 as shown below:

[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]

Thus, we infer this is a double replacement reaction due to the exchange of both cations, K and Pb with both anions, I and NO3. Moreover, we can tell this balanced reaction is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).

Regards!

Answer:

0.000731 grams aluminium sulfate

46.0 mols ammonia

Explanation:

ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol

[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]

NH3 has a molar mass of 17.031 g/mol

[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]

Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]

we have to find the 0.250 moles of aluminum sulphate.

[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]

[tex]\\\\\\[/tex]

Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]

We have to find 2.70 grams of ammonia

[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]

Answer:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Explanation:

Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:

CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)

Answer:

[tex]\boxed {\boxed {\sf 63.0 \%}}[/tex]

Explanation:

The percent yield is the ratio of the actual yield to the theoretical yield.

[tex]percent \ yield = \frac{actual \ yield}{theoretical \ yield} * 100[/tex]

We can substitute the known values into the formula.

[tex]percent \ yield= \frac{ 14.8 \ g}{23.5 \ g}*100[/tex]

Divide.

[tex]percent \ yield = 0.629787234043*100[/tex]

Multiply.

[tex]percent \ yield = 62.9787234043[/tex]

The original measurements for the theoretical and actual yields have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place.

The 7 to the right, in the hundredths place, tells us to round the 9 up to a 0. Since we rounded up to 0, we have to move to the next place to the left and round the 2 up to a 3.

[tex]percent \ yield \approx 63.0[/tex]

The percent yield is approximately 63.0 percent.

Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders

Answer:

irreversible.

I hope this will help you

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

Explanation:

Answer:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Explanation:

The strongest acid is the one that has the higher Ka. Now, the weakest conjugate base is the conjugate base of the strongest acid and vice versa:

In the problem, we have only conjugate bases, as the HCN is the weakest acid, the strongest conjugate base is NaCN, then KClO and as last C2H5NH3Br and NaI (The conjugate base of a strong acid, HI).

The strongest base has the higher pH, that means. Thus, the rank in order of increasing pH is:

0.1 M Nal

0.1M C2H5NH3Br

0.1M KClO

0.1M NaCN

Answer:

a

Explanation:

Beer-Lambert Law shows the relationship between the factors affecting the absorbance of a sample in relation to the concentration. These factors are:

the concentration c, path length (l), and the molar absorptivity (ε).

As a result, more radiation is assimilated as the concentration rises, and the absorbance rises as well. However, the longer the path length, the increase in the number of molecules and the higher the absorbance.

Thus, the straight-line equation for Beer-Lambert's law is:

A = εcl

From the above explanation, the option that doesn't relate to the deviations from linearity of Beer's law plot is in Option (a).

Answer: 0.085 (Ms)⁻¹

Explanation: Half life = 12 s

is the initial concentration = 0.98 M

Half life expression for second order kinetic is:

k = 0.085 (Ms)⁻¹

The rate constant for this reaction is 0.085 (Ms)⁻¹ .

Answer:

2

Explanation:

Lead(|V) fluoride

Ammonium Nitrate

Lithium sulfide

For the rules, I don't know what you were taught. I just do it intuitively since I have done so much chemistry.

The first one the roman numerals represents the charge of the lead which much match the 4- charge from the 4 fluorides.

The second one is just two polyatomic ions which you just have to remember.

The last one is the typical ionic compound naming technique i guess.

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

Answer:

Explanation:

a) Ionic

Lithium oxide

b) Covalent

[tex]$\ce{N_2O_3}$[/tex]

c) Covalent

Phosphorus trichloride

d) Ionic

[tex]Mn_2O_3[/tex]