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anyanavicka [17]

2 years ago

11

A radioisotope is placed near a radiation detector, which registers 80 counts per second. Eight hours later, the detector regist

ers 5 counts per second. What is the half- life of the radioactive isotope

1 answer:

sukhopar [10]2 years ago

8 0

The half life is that the material loses 1/2 of its mass every two hours

You might be interested in

What types of compounds are CaCl2, Cu, C2H6, respectively.

mina [271]

Answer:

Ionic, metal, organic

Explanation:

In this case, we have to analyze each compound:

-) $CaCl_2$

In this compound, we have a non-metal atom (Cl) and a metal atom (Ca) . So, we will have a high electronegativity difference between these atoms, With this in mind, we will have an ionic bond. Ions can be produced:

$CaCl_2~->~Ca^+^2~+~2Cl^-$

The cation would be $Ca^+^2$ and the anion is $Cl^-$ . So, we will have an <u>ionic compound.</u>

-) $Cu$

In this case, we have a single atom. If we check the periodic table we will find this atom in the transition metals section (in the middle of the periodic table). So, this indicates that Cu (Copper) is a <u>metal.</u>

-) $C_2H_6$

In this molecule, we have single bonds between carbon and hydrogen. The electronegativity difference between C and H are not high enough to produce ions. So, with this in mind, we will have covalent bonds. This is the main characteristic of <u>organic compounds. </u> (See figure 1)

5 0

2 years ago

Avanti works in a bookstore. She has four books and is going to place them in two stacks. The diagram above shows the books befo

Feliz [49]

Answer:

Well they didn't transfer any energy when they weren’t touching and it did t produce any energy if it didn’t move. Since they are on top of each other they are causing momentum on each other creating kinetic energy

Explanation:

3 0

1 year ago

A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =

IgorC [24]

According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
= 2.16 x10^2

8 0

2 years ago

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

What is the benefit of having a limiting reagent when performing a lab experiment

Shalnov [3]

Answer : The role of limiting reagent or reactant is important in a chemical reaction because it can help the chemist to predict that complete amount of reactant is consumed, as it is limiting the reaction, only required moles of products can get formed instead of the theoretical yield where the perfect amount is used.

In short, Limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is found to be complete.

3 0

2 years ago

Read 2 more answers