The half-life of phosphorous-32 is 14.28 days. How many days are required for 4.68 g to remain if you start with 74.88
grams

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

[tex]\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})[/tex]

Where:

Q: is the energy transferred = 5.0 MJ

[tex]k_{B}[/tex]: is the Boltzmann constant = 1.38x10⁻²³ J/K  

[tex]T_{i}[/tex]: is the initial temperature = 1000 K

[tex]T_{f}[/tex]: is the final temperature = 500 K

Hence, the entropy change is:

[tex] \Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26} [/tex]                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

[tex]\huge\mathsf{\red{\underline{\underline{Compound}}}}[/tex]

[tex]{\green{\dashrightarrow}}[/tex]A chemical compound is a chemical substance that is made of two or more atoms of different elements that share a chemical bond.

[tex]{\green{\dashrightarrow}}[/tex]A chemical formula represents the ratio of atoms per element that make up the chemical compound.

[tex]\large{\pink{\sf{5~ Examples~ of~ Compound~ are:-}}}[/tex]

Molar mass of CH2NH2COOH - 75

Given mass of CH2NH2COOH - 30

Moles of CH2NH2COOH = Given mass/ Molar mass

moles of CH2NH2COOH = 30/75 = 0.4 mol

One mole of CH2NH2COOH contains 32 gram of oxygen

0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen

Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!

The mass of oxygen in 30 g of CH₂NH₂COOH is 12.8 g.

We'll begin by calculating the mass of 1 mole of CH₂NH₂COOH and the mass of oxygen in the compound.

1 mole of CH₂NH₂COOH = 12 + (2×1) + 14 + (2×1) + 12 + 16 + 16 + 1 = 75 g

Mass of oxygen in CH₂NH₂COOH = 16 + 16 = 32 g

Thus,

75 g of CH₂NH₂COOH contains 32 g of Oxygen.

With the above information, we can obtain the mass of oxygen in 30 g of CH₂NH₂COOH. This can be obtained as follow:

75 g of CH₂NH₂COOH contains 32 g of Oxygen.

Therefore,

30 g of CH₂NH₂COOH will contain = (30 × 32) / 75 = 12.8 g of Oxygen.

Thus, 12.8 g of Oxygen is present in 30 g of CH₂NH₂COOH

Calculate the mass of oxygen in 30 g of CH₂NH₂COOH.

Learn more about mass composition:

https://brainly.com/question/11617445

Answer:

Dispersion forces

that is the answer

Answer:

[tex](NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)[/tex]

Explanation:

Hello there!

In this case, according to the given information, we can set up the appropriate chemical equation when ammonium phosphate reacts with sodium chloride in aqueous solution:

[tex](NH_4)_3PO_4(aq)+NaCl(aq)\rightarrow NH_4Cl(aq)+Na_3PO_4(aq)[/tex]

Which stands for a double replacement reaction, whereby ammonium changes phosphate to chloride and sodium changes chloride to phosphate on the products side. In addition, we can balance the aforementioned equation as shown below:

[tex](NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)[/tex]

Regards!

Answer:

From the diagram above...

its explicitly show a compound that's dissociating or Decomposing into its Constituents.

Answer ....OPTION C.

Answer:1.12 g

Explanation:Molar mass of compound ​FeSO4(NH4)2SO4.6H2O = 392.14 g mol-1

Number of moles of compound in 7.84g sample = 7.84 g/ 392.14 g mol-1

                                                                       = 0.02 mole

1 mole of sample contain 1 mole of Fe

So, 0.02 mole of sample would contain 0.02 moles of Fe.

Mass of Fe equivalent to 0.02 mole = 0.02 mol x molar mass of Fe

                                                    = 0.02 mol x 55.84 g mol-1

                                                   = 1.12 g

Answer:

methyl ethanoate

Explanation:

To name the compound given above, the following must be obtained:

1. Determine the functional group of the compound.

In this case, the functional group is R–COOR' where R and R' are alkyl groups. Thus, the compound is an ester.

2. Determine the longest chain before the functional group and the compound after the functional group.

In this case, the longest chain before the functional group is carbon 2 i.e ethane and the compound after the functional group is methyl.

3. Name the compound by naming the compound after the functional group first, followed by the compound before the functional group and ending it with –oate.

This is illustrated below:

After the functional group => methyl

Before the functional group => ethane

Name of the compound => methyl ethanoate

Answer:

B. Dividing the number by 1,000

Explanation:

Every time you move the decimal ,add a zero.Since the decimal is moved three times ,there are three zeroes. When you move left ,the number is getting smaller so it is division.

Answer:

the answer is copper wire

Answer:

Fusion has powered the sun for billions of years.Yet despite decades of effort, scientists and engineers have been unable to generate sustained nuclear fusion here on Earth. In fact, it's long been joked that fusion is 50 years away, and will always be.

Answer:

B. Ultraviolet

Explanation:

UV has a higher frequency and shorter wavelength than visible light

Answer:

Mass of solute in Kg = 0.07 Kg

Explanation:

Given

Molecular weight of the chemical = 28 grams per mole

Solution consists of 2.5 moles of solution

Mass of solute in the solution = 2.5 *28 = 70 grams

Mass of solute in Kg =  70 grams/1000 = 0.07 Kg

Answer:

the chemical name for soap is a lot, for example there is Sodium Talowate, Sodium Palmate, and even Sodium Cocoate, but there is more

Explanation:

the chemical formula for it is RCOO-Na+ it has 12 - 18 carbon atoms.

Answer:

D0wnload Phot0Math......................

Explanation:

Answer:

~69.744 moles of Ca

Explanation:

Using Avogadro's constant , we know that:

1 mole = 6.022 x 10^23 atoms

S0, the number of moles in 4.20 x 10^25 atoms of Ca:

=(4.20 x 10^25 x 1 )/(6.022 x 10^23)

~69.744 moles of Ca

Q2:How many atoms are in 0.35 moles of oxygen?

1 mole = 6.022 x 10^23 atoms

S0, the number of atoms in 0.35 moles of  oxygen:

=[0.35 x (6.022 x 10^23)]

=2.1077 x 10^23 atoms of Oxygen

Hope it helps:)

The question is incomplete, the complete question is;

Which produces the greatest number of ions when one mole dissolves in water? a. NaCl b. NH4Cl c. NH4NO3 d. Na2SO4

Answer:

Na2SO4

Explanation:

If we consider the compounds listed in the options one after the other;

NaCl produces two moles of ions in solution

NH4Cl produces two moles of ions in solution

NH4NO3  produces two moles of ions in solution

Na2SO4 produces three moles of ions in solution

We can see that Na2SO4 produces the greatest number of ions when one mole of the substance  is dissolved in water, hence the answer above.

Answer:

Linear

Explanation:

Answer:first part is A and C, second part is C, third part is 40 m/s and the fourth part is zero

Explanation: I guessed and got them right lol

Answer:

5.56g of FeCl2 can be produced

Explanation:

To solve this question we must find the moles of each reactant. With the moles and the chemical equation we can find limiting reactant. With limiting reactant we can find the moles of FeCl2 and its mass as follows:

Moles HCl:

1.5L * (0.25mol / L) = 0.375 moles HCl

Moles Fe -Molar mass: 55.845g/mol-

2.45g * (1mol / 55.845g) = 0.0439 moles Fe

For a complete reaction of 0.375 moles HCl are needed:

0.375 moles HCl * (1mol Fe / 2mol HCl) = 0.1875 moles Fe

As there are just 0.0439 moles Fe, Fe is limiting reactant

1mol of Fe produce 1 mole of FeCl2, 0.0439 moles Fe produce 0.0439 moles of FeCl2. The mass is:

Mass FeCl2 -Molar mass: 126.751g/mol:

0.0439 moles Fe * (126.751g / mol) =

Answer:

The correct answer is "142 g".

Explanation:

According to the question,

⇒ [tex]Cu^2+(aq)+2e^ ------> Cu(s)[/tex]

where,

Z = 2

F = 96500c

C = 30 A

t = 4h

 = [tex]4\times 60\times 60[/tex]

 = [tex]14400 \ sec[/tex]

We know that,

Atomic mass of Cu(M) = 63.5 g/mole

Now,

⇒ [tex]W=\frac{Mct}{ZF}[/tex]

On putting the values, we get

        [tex]=\frac{63.5\times 30\times 14400}{2\times 96500}[/tex]

        [tex]=\frac{27432000}{193000}[/tex]

        [tex]=142 \ g[/tex]

Answer:

The molality is "6.86 m".

Explanation:

Given:

Mass of of HCL,

= 25 g

Mass of water,

= 100 g

Density,

= 1.19 g/mL

Total mass of solution,

= 125 g

Now,

The number of moles of HCl will be:

= [tex]\frac{25}{36.458}[/tex]

= [tex]0.6857[/tex]

The solution volume will be:

= [tex]\frac{125}{1.19}[/tex]

= [tex]105.04 \ mL[/tex]

hence,

The molality will be:

= [tex]\frac{No. \ of \ moles \ of \ solute}{Mass \ of \ solvent (Kg)}[/tex]

= [tex]\frac{0.6857}{0.1}[/tex]

= [tex]6.857 \ m[/tex]

or,

= [tex]6.86 \ m[/tex]

Answer:

negatively

Explanation:

when soil erosion happens,by the agent of water for example,the soil will eventually lose it's valuable minerals and become infertile, people won't be able to grow crops there.

Answer:

they have the same molar solubility at 44.5 degrees celsius

Answer:

Green

Explanation:

pH : < 3 = Red

pH : 3 - 6 = Orange or yellow

pH : 7 = Green

pH : 8 - 11 = Blue

Answer:

6967 years

Explanation:

The radioactive substance left after a periodical t year can be expressed by using the formula:

[tex]Q(t) = Q_oe^{-kt}[/tex]

here;

[tex]Q_o[/tex] = the radioactive initial value.

We need to understand that provided that the radioactive substance will get reduced to half of the provided initial amount  after a periodic time, Then:

the half-life of the radioactive substance left is:

[tex]Q(h) = \dfrac{Q_o}{2}[/tex]

Given that:

the half-life = 1500 years

[tex]\dfrac{Q_o}{2}= Q_o e^{-k\times 1500} \\ \\ \dfrac{Q_o}{2}= Q_o e^{ -1500k}[/tex]

Divide both sides by [tex]Q_o[/tex]

[tex]\dfrac{1}{2} =e^{-1500k}[/tex]

Then, find the natural logarithm of both side;

[tex]\mathtt{In \dfrac{1}{2} = -1500 k}[/tex]

[tex]k = \dfrac{1}{-1500}\mathtt{In}\dfrac{1}{2}[/tex]

k = 0.000462

So, after a particular (t) time, a 250 kg radium sample was reduced to 10 kg;

Then:

[tex]10 = 250 e^{-0.000462t}[/tex]

[tex]0.04 = e^{-0.000462t}[/tex]

From both sides, finding the natural logarithm, we have:

In(0.04) = -0.000462t

[tex]t = \dfrac{In(0.04)}{-0.000462}[/tex]

t = 6967.26

Thus, it will take approximately 6967 years for a 250 kg radium sample to get reduced to 10 kg.

Answer:

The explanation according to the given question is summarized below.

Explanation:

Given:

Tris,

= 250 mM

Glycine,

= 1.92 M

According to the solution,

For the dilution pf 10X to 1X buffer, we get

= [tex]1 \ ml \ of \ 10X \ buffer +9 \ ml \ of \ distilled \ water[/tex]

= [tex]10[/tex]

i.e.,

⇒ [tex]10X \ to \ 1X=1:10 \ dilution[/tex]

Now,

⇒ [tex]10X (250 \ mM\ Tris \ HCl, 1.92M\ Glycine, and\ 1 \ percent (\frac{w}{v} ) SDS) ---->1X(25 \ mM \ Tris \ HCl,0.193 M\ Glycine, and \ 0.1 \ percent(\frac{w}{v} )SDS)[/tex]

Answer:

HNO3 is a potent acid, a base, a nitrating agent and a heavy oxidising agent at times. In the presence of a stronger acid, it serves as a base.

Explanation:

Answer:

It is a strong acid

Explanation:

Answer:

The answer is "[tex]\bold{21.616\ \ \frac{J}{mol K}}[/tex]".

Explanation:

when it is a constant pressure:

[tex]Cp= \frac{\Delta q}{n\Delta T} \\\\[/tex]

[tex]=\frac{229}{3\times 2.55}\\\\=\frac{229}{7.65}\\\\=29.93 \frac{J}{mol\ K}\\[/tex]

and,  

[tex]Cp-Cv=R\ \ (Mayer's\ formula)[/tex]

then,

[tex]Cv=Cp-R\\\\[/tex]  

    [tex]=29.93-8.314\\\\=21.616\ \ \frac{J}{mol K}[/tex]