Please help! Thanks in advance!

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

Answer:

Keep temperature constant and increase the pressure of the reaction. The rate of reaction increases.

Explanation:

First of all, the question is asking us to design an experiment to investigate the effect of pressure on the rate of reaction hence the pressure can not be held constant since it is the variable under investigation. This eliminates the first option.

Secondly, increasing the pressure of the reaction means that particles of the gas collide more frequently leading to a greater number of effective collisions and a consequent increase in the rate of reaction according to the collision theory.

Hence the answer above.

Answer: The mass of P-32 left from the original sample is 32.07 mg

Explanation:

All radioactive decay processes follow first-order reactions.

Calculating rate constant for first order reaction using half life:

[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)

[tex]t_{1/2}[/tex] = half life period = 14.3 days

k = rate constant = ?

Putting values in equation 1:

[tex]k=\frac{0.693}{14.3days}\\\\k=0.0485days^{-1}[/tex]

The integrated rate law equation for first-order kinetics:

[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(2)

Given values:

a = initial concentration of reactant = 175 mg

a - x = concentration of reactant left after time 't' = ? mg

t = time period = 35.0 days

Putting values in equation 2:

[tex]0.0485days^{-1}=\frac{2.303}{35.0 days}\log (\frac{175}{a-x})\\\\\log (\frac{175}{a-x})=\frac{0.0485\times 35.0}{2.303}\\\\\log (\frac{175}{a-x})=0.737\\\\\frac{175}{a-x}=10^{0.737}\\\\a-x=\frac{175}{5.457}=32.07mg[/tex]

Hence, the mass of P-32 left from the original sample is 32.07 mg

Answer:

D

Explanation:

Answer:

Mean

Explanation:

The mean of a series of measurements is calculated when all the measurements are added up and then divided by the number of measurements taken, as follows:

As there are 4 measurements, the mean is:

Answer:

A

Explanation:

Answer:

Explanation:

NaL

Difluorine - F2

Molar Mass Bond Polarity F₂ Fluorine Gas Fluorine

Products

Sodium Fluoride - NaF

Molar Mass Bond Polarity Oxidation State Floridine Sodium Monofluoride Naf

L2

Answer:

t(9.32% remaining) = 203 yrs (3 sig. figs.)

Explanation:

All radioactive decay follows a 1st order decay profile. This is defined by the expression ...

A =A₀e^-k·t

A = final activity = 9.32%

A₀ = initial activity = 100%

e = base of natural logs

k = rate constant = 0.693/t(1/2) = (0.693/30.2) yrs⁻¹ = 0.023 yrs⁻¹

t = time of decay = ln(A/A₀)/-k = ln(9.32%/100%)/-0.023 yrs⁻¹

 = 203.286637 yrs (calc. ans.)

 ≅ 203 yrs (3 sig. figs.)

Answer:

If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to reduce a chemical species while simultaneously forming iodine.

Iodine always used in a solution excess KI is given to aid in the solubilization of free iodine, which would be insoluble in clean water during normal circumstances.

Iodine is a kind of element which are mainly used in iodometry titration. It can be represented by I.

A solution would be a homogenous mixture of two components, usually a solute as well as a solvent.

Iodimetry would be a technique that uses a standard iodine solution as a titrant for such an oxidizable analyte. When an excessive amount of iodide is used to decrease a chemical while somehow producing iodine.

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Answer: The mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.

Explanation:

The given balanced reaction equation is as follows.

[tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex]

Here, the mole ration of Al and hydrogen produced is 2 : 3

As mass of aluminum is given as 26.98 g. So, moles of aluminum (molar mass = 26.98 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{26.98 g}{26.98 g/mol}\\= 1 mol[/tex]

So, when 1 mole of Al reacted then 1.5 moles of hydrogen is produced as per the given mole ratio.

Therefore, mass of hydrogen formed is calculated as follows.

[tex]mass = moles \times molar mass\\= 1.5 mol \times 2.02 g/mol\\= 3.03 g[/tex]

Thus, we can conclude that the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.

Answer:

Number of mole in H2O = 5.4 moles

Explanation:

Given:

Mass of H2O = 97.3 gram

Find:

Number of mole in H2O

Computation:

We know that molar mass of H2O = 18 g/mole

So,

Number of mole = Given Mass / Molar mass

Number of mole in H2O = Mass of H2O / Molar mass of H2O

Number of mole in H2O = 97.3 / 18

Number of mole in H2O = 5.4055

Number of mole in H2O = 5.4 moles

Explanation:

An atom consists of two regions. The first is the tiny atomic nucleus, which is in the center of the atom and contains positively charged particles called protons and neutral, uncharged, particles called neutrons. ... Most atoms contain all three of these types of subatomic particles—protons, electrons, and neutrons.

Answer:

897154.2 J

Explanation:

Applying,

q = mcΔT.............. Equation 1

Where q = amount of energy released, m = mass of water, c = specific heat capacity of water, ΔT = Change in temperature

From the question,

Given: m = 2.25 kg = 2250 g, c = 4.184 J/g°C, ΔT = (4.2-99.5) = -95.3°C

Substitute these values into equation 1

q = 2250(4.184)(-95.3)

q = -897154.2 J

q =

Hence the amount of heat released is 897154.2 J

Answer: The equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]

Explanation:

The equilibrium constant is defined as the ratio of the concentration of the products to the concentration of reactants each raised to the power of their stoichiometric coefficients.

The concentration of all the solids and liquids are considered to be 1 in the expression of equilibrium constant

For the given chemical equation:

[tex]2KClO_3(s)\rightleftharpoons 2KCl(s)+3O_2(aq)[/tex]

The expression of equilibrium constant follows:

[tex]K_{eq}=[O_2]^3[/tex]

Hence, the equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]

Answer:

The iron core, copper wire, and an electricity source.

Explanation: Me

Answer:

The given molecules are:

a. C6H13NH2

b. CH3OH

c. CH4

d. C5H11OH

e. CO2

Explanation:

The hydrogen bond is a weak electrostatic force of attarction that exists between covalently bonded hydrogen (of -OH or -NH2 or HF) with a highly electronegative atom like N,O and F.

Hydrogen bonding is of two types:

Intermolecular hydrogen bond (exists between two molecules)

Intramolecular hydrogen bond(exists within a molecule).

For example intermolecular hydrogen bond in water is shown below:

Among the given options,

a. C6H13NH2 has -NH2 linkage which leads to hydrogen bond formation.

b. CH3OH has -OH bond and it leads to hydrogen bond fomation.

d. C5H11OH has also -OH bond and it leads to hydrogen bond fomation.

Reamining molecules, CH4 and CO2 do not form hydrogen bond.

Hence, answer is:

options a,b,d.

Answer:

the answer is 3

Explanation:

hope this works

Answer:

Please find the complete question in the attached file.

Explanation:

It would only be radioactive if the DNA molecule that employed the poly-T rand as templates. Its other molecule of the daughter would not have been radioactive as it did not need dATP for its replication. While each strand of the second molecule includes t, simultaneous reproduction dATP from both daughter molecules is needed so that each of those is radioactive.

Answer:

i think it's variable pressure

if not soo advance sorry :)

Answer:

[tex]m_{Al_2O_3}=118.27gAl_2O_3[/tex]

Explanation:

Hello there!

In this case, if we consider the following chemical reaction, whereby Al2O3 is produced from Al and FeO:

[tex]3FeO+2Al\rightarrow 3Fe+Al_2O_3[/tex]

Thus, since there is 3:1 mole ratio of FeO to Al2O3, it turns out feasible for us to use their molar masses, 71.844 g/mol and 101.96 g/mol respectively, to obtain the grams of the latter as follows:

[tex]m_{Al_2O_3}=250.gFeO*\frac{1molFeO}{71.844gFeO}*\frac{1molAl_2O_3}{3molFeO} *\frac{101.96gAl_2O_3}{1molAl_2O_3}\\\\m_{Al_2O_3}=118.27gAl_2O_3[/tex]

Regards!

Answer:

Explanation:

Speed of light = Frequency * Wavelength

Speed of light is 3*10^8 m/s

Wavelength = Speed / Frequency = 3*10^8 / 3.7*10^14

= 8.11*10^(-7) m

= 811 nanometer

The wavelength of light will be

The wavelength of such a periodic wave would be the interval in which the shape of the wave repeats.

Calculation of wavelength of light

It is given that,

frequency = 3.7 x [tex]10^{14}[/tex] Hz

Speed of light = 3 × [tex]10^{8}[/tex] m/s

frequency × wavelength = speed of light.

wavelength =  speed of light / frequency

wavelength =  3 × [tex]10^{8}[/tex] m/s / 3.7 x [tex]10^{14}[/tex] Hz

wavelength = 0.81 × [tex]10^{-7}[/tex] m

wavelength = 8.1 × [tex]10^{-7}[/tex]

Therefore, wavelength of light will be  8.1 × [tex]10^{-7}[/tex].

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Answer:

CH3CH2CH3

Explanation:

Dipole moment is the measure of the polarity of a chemical bond. It is the extent of charge separation in a molecule.

Dipole moment is the product of the magnitude of charge and the distance separating the charges from each other.

The molecule having the lowest dipole moment among the options is the molecule that has the least polarity. The least polar molecule among the options is CH3CH2CH3, it has no polar bonds in its structure.

We have that for the Question "Identify the compound with the lowest dipole moment. Identify the compound with the lowest dipole moment. CH3CH2CH3 CH3OCH3 CH3CHO CH3OH CH3CN "

it can be said that

 From the question we are given

CH3CH2CH3

CH3OCH3

CH3CHO

CH3OH

CH3CN

Generally alkanes have the lowest dipole moment, they have C-H bond which are non polar.

Therefore,

[tex]CH_3CH_2CH_3[/tex] have the lowest dipole moment

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Answer:

a

Explanation:

a

Answer:

B. Isotopes of the element.

Explanation:

Isotopes are basically atoms of the same element that contain the same number of protons, but a different number of neutrons.

Answer:

6.85 g

Explanation:

Step 1: Given data

Step 2: Calculate the moles of sucrose (solute) required

Molarity is equal to the moles of solute divided by the liters of solution.

M = moles of solute / liters of solution

moles of solute = M × liters of solution

moles of solute = 0.200 mol/L × 0.10000 L = 0.0200 mol

Step 3: Calculate the mass corresponding to 0.0200 moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

0.0200 mol × 342.3 g/mol = 6.85 g

The amount of sucrose (in grams) must be weighed out for the given reaction is 6.846g.

Mass of any substance will be calculated by using the moles as:
n = W/M, where

W = required mass

M = molar mass

Given that, molarity of sucrose = 0.2 M

Volume of solution = 100mL = 0.1 L

Relation between moles and molarity is represented as:
M = n/V

On putting values on the above equation we get,

n = (0.2)(0.1) = 0.02 moles

We know that molar mass of sucrose (C₁₂H₂₂O₁₁) = 342.3 g/mole

Now we calculate the required mass by putting values on the first equation as:
W = (0.02)(342.3) = 6.846g

Hence, the required mass of sucrose is 6.846g.

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Answer:

11.1 cm

Explanation:

First we convert 58.1 pounds into grams:

Then we calculate the volume of the gold cube, using the given density:

26353.69 g ÷ 19.3 g/mL = 1365.48 mL

With the volume of a cube we can calculate the side length:

Answer:

2.00 × 10⁻³ g

Explanation:

Step 1: Write the balanced decomposition reaction

2 NaHCO₃ ⇒ Na₂CO₃ + CO₂ + H₂O

Step 2: Calculate the moles corresponding to 0.0118 g of Na₂CO₃

The molar mass of Na₂CO₃ is 105.99 g/mol.

0.0118 g × 1 mol/105.99 g = 1.11 × 10⁻⁴ mol

Step 3: Calculate the moles of H₂O produced with 1.11 × 10⁻⁴ moles of Na₂CO₃

The molar ratio of Na₂CO₃ to H₂O is 1:1. The moles of H₂O produced are 1/1 × 1.11 × 10⁻⁴ mol = 1.11 × 10⁻⁴ mol.

Step 4: Calculate the mass corresponding to 1.11 × 10⁻⁴ moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

1.11 × 10⁻⁴ mol × 18.02 g/mol =  2.00 × 10⁻³ g

Answer:

D. The planet that is closer to the Sun

Answer: