Organize the following metals from least metallic to most metallic :barium,zinc,iron, and strontium

Determine the empirical formula of a

k0ka [10]

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold = 2.6444g

Mass of Chlorine = 0.476g

Unknown:

Empirical formula = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements Au Cl

Mass 2.6444 0.476

Molar mass 197 35.5

Number of moles 2.6444/197 0.476/35.5

0.013 0.013

Divide by the

smallest 0.013/0.013 0.013/0.013

1 1

The empirical formula of the compound is AuCl

3 0

2 years ago

which of the following pairs of elements could react to form an ionic compound? A. potassium and calcium B. sulfur and oxygen C.

Grace [21]

The correct option is FLUORINE AND COPPER.
An ionic compound is usually formed by the combination of a metal and a non metal, the metal usually act as an electron donor while the non metal act as an electron acceptor. Thus, in ionic compounds, there is total transfer of electrons from the metal to the non metal. In the question given here, copper is the metal while the fluorine is the non metal.<span />

4 0

3 years ago

The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many ye

ryzh [129]

Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

$ln \frac{N}{N_0} = - k*t$ ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

$T _\frac{1}{2} = \frac{ln2}{k}$ ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of $T_\frac{1}{2}$ in above formula (equation 2) :

$28.8 yrs = \frac{ln 2}{ k }$

Multiply both side by k

$28.8 yrs * k = \frac{ln 2 }{k} * k$

Dividing both side by 28.8 yrs

$\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}$

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

$ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t$

$ln (0.0971 ) = -0.0241 yrs ^-^1 * t$

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

$\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}$

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

3 0

3 years ago

Read 2 more answers

How many hydrogen atoms are in 5.80 mol of ammonium sulfide

IRINA_888 [86]

Moles of ammonium sulfide = 5.80 mol

The formula of ammonium sulfide is (NH₄)₂S

So each molecule of ammonium sulfide has (4 x 2) or 8 atoms of H

One mole of ammonium sulfide has 8 moles of H

5.80 mol of ammonium sulfide has (8 x 5.8) or 46.4 moles of H

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ atoms

46.4 moles of H x (6.022 x 10²³ atoms/ 1 mole of H)

= 2.8 x 10²⁵ H atoms

Therefore, 2.8 x 10²⁵ H atoms are in 5.80 mol of ammonium sulfide.

5 0

3 years ago

In the name carbon dioxide, the prefix of the second word indicates that a molecule of carbon dioxide contains

xz_007 [3.2K]

the prefix of the second word indicates that a molecule of carbon dioxide indicates that the compound contains two oxygen atoms

8 0

3 years ago

Read 2 more answers