A compound has an empirical formula of

A small amount of a solid is added to water. The observation made after fifteen minutes is shown in the figure. Which of these s

natulia [17]

Answer:

B) sand

Explanation:

A) Oil is a wrong choice because it is a liquid not a solid and also if it is oil, it will float over the water surface as a droplets.

B) Sand is the right choice, because sand is a solid and it does not dissolve in water and stabilizes at the bottom.

C) Sugar is a wrong choice, because small amount of sugar will dissolve in water and be a homogeneous solution and does not appear as a particles.

D) Wood ships is also a wrong choice, even it is a solid and does not dissolve in water, but it will float over the water surface.

4 0

3 years ago

Determine the fraction of an original sample of tritium that remains unchanged after 24.62 years.

patriot [66]

Answer: The answer is 1/4

Explanation:

7 0

3 years ago

What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

Round

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.4184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately 147 grams.

3 0

3 years ago

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

viktelen [127]

Answer : The concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is $SCN^-$ and $Fe^{3+}$ is excess reagent.