All categories

2 years ago

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

Chemistry

1 answer:

Charra [1.4K]2 years ago

3 0

Answer:

heres the link of the answer

boldniwally.com

You might be interested in

If you have 1.005 x 1024 molecules of Ca(OH)2 how many atoms do you have? Moles? Grams?

ira [324]

Answer:

5.025x10²⁴ atoms

1.6686 moles

123.63 g

Explanation:

In one Ca(OH)₂ molecule, there are 5 atoms (1 Ca, 2 O and 2 H). This means that in 1.005x10²⁴ Ca(OH)₂ molecules, there are (5 * 1.005x10²⁴) 5.025x10²⁴ atoms.

To convert molecules into moles, we use Avogadro's number:

1.005x10²⁴ molecules * $\frac{1mol}{6.023*10^{23}molecules}$ = 1.6686 moles

Finally, we convert moles into grams using the molar mass of Ca(OH)₂:

1.6686 moles * 74.093 g/mol = 123.63 g

4 0

3 years ago

What human life process requires hydrochloric acid? THANKS!!!

notka56 [123]

The answer is Enzyme

7 0

3 years ago

What is the half-life of 20 g of a radioactive sample if 5 g remain after 8 minutes?

Anni [7]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 20 g

m (final mass after time T) = 5 g

x (number of periods elapsed) = ?

P (Half-life) = ? (in minutes)

T (Elapsed time for sample reduction) = 8 minutes

Let's find the number of periods elapsed (x), let us see:

$m = \dfrac{m_o}{2^x}$

$5 = \dfrac{20}{2^x}$

$2^x = \dfrac{20}{5}$

$2^x = 4$

$2^x = 2^2$

$\boxed{x = 2}$

Now, let's find the half-life (P) of the radioactive sample, let's see:

$T = x*P$

$8 = 2*P$

$2\:P = 8$

$P = \dfrac{8}{2}$

$\boxed{\boxed{P = 4\:minutes}}\Longleftarrow(Half-Life)\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

5 0

2 years ago

Hi pls help pls pls. this is due at 11:59 help

Sveta_85 [38]

Answer:

mass of salt 6.72g

Explanation:

mass of the combine mass with salt- mass of the empty beaker

36.176g−29.456g=6.72g

7 0

1 year ago

Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction

11Alexandr11 [23.1K]

Answer:

Balanced equation:

$Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)$

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

$Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate$

$Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)$

Ionic equation:

$Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}$

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

$Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)$

4 0

2 years ago