As you move left to right across a period, which of the following always increases?

Pleaseeeeeee hlepThe half-life of cobalt-60 is 5.26 years. After 10.52 years, 5 grams of a 20-gram sample will remain.

vladimir1956 [14]

The half-life of cobalt-60 is 5.26 years. After 10.52 years, 5 grams of a 20-gram sample will remain is TRUE

Explanation:

Mass of cobalt = 20 g

Half-life = 5.26 years

Mass remains after 10.52 years = 5 g

This can be solved by using given below formula, $m(t)=m_{o}\left(\frac{1}{2}\right)^{\frac{l}{5.26}}$

$m_{0}$ = initial mass

t = number of years from when the mass was m_0

m(t) = remaining mass after t years

Number of half-lives = $\frac{\text { Time elapsed }}{\text { Half -life }}$

Number of half-lives = $\frac{10.52 \text { years }}{5.26 \text { years }}$

Number of half-lives = 2

At time zero = 20 g

At first half-life = $\frac{20\ g}{2}$ = 10 g

At second half life = $\frac{10\g}{2}$ = 5 g

The given statement is true.

4 0

2 years ago

Make a quantitative and qualitative statement about the following:. water, carbon, iron, hydrogen gas, sucrose, table salt, merc

gladu [14]

The elements in this list are mercury, gold, iron, carbon and hydrogen. The compounds in this list, on the other hand, are sucrose, table salt, water and air. Elements are composed only of one substance while compounds are composed of two or more substances.

8 0

2 years ago

Read 2 more answers

A 0.04328 g sample of gas occupies 10.0-mL at 294.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C

stepan [7]

Answer:

The molecular formula of the gas sample = $C_2Cl_2$

Lewis structure is shown in the image below.

The geometry around each carbon atom is linear.

Explanation:

Given that:

Temperature = 294.0 K

V = 10.0 mL = 0.01 L ( 1 mL = 0.001 L )

Pressure = 1.10 atm

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

1.10 atm × 0.01 L = n ×0.0821 L atm/ K mol × 294.0 K

⇒n = 0.0004557 mol

Given, mass = 0.04328 g

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0004557\ mole= \frac{0.04328\ g}{Molar\ mass}$

Molar mass of the gas sample = 94.9747 g/mol

Given that:-

% of C = 25.305

Molar mass of C = 12.0107 g/mol

% moles of C = 25.305 / 12.0107 = 2.1069

% of Cl = 74.695

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 74.695 / 35.453 = 2.1069

Taking the simplest ratio for C and Cl as:

2.1069 : 2.1069 = 1 : 1

The empirical formula is = $CCl$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 35.5 = 47.5 g/mol

Molar mass = 94.9747 g/mol

So,

Molecular mass = n × Empirical mass

94.9747 = n × 47.5

⇒ n = 2

The molecular formula of the gas sample = $C_2Cl_2$

Also,

Valence electrons of carbon = 4

Valence electrons of Chlorine = 7

The total number of the valence electrons = 4*2 + 7*2 = 22

The Lewis structure is drawn in such a way that the octet of each atom in the molecule is complete. So,

The Lewis structure is shown in the image below.

According to the theory, the atoms will form a geometry in such a way that there is minimum repulsion and maximum stability. The carbon atoms are sp hybridized.

So, it is of linear shape.

3 0

3 years ago

How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo

riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$