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ohaa [14]
3 years ago
15

Convert 19,864 mm to m​

Chemistry
2 answers:
navik [9.2K]3 years ago
3 0
19.864 meters would be the correct answer
mart [117]3 years ago
3 0

I think 19,864 mm converted to m is 19.864 I'm not 100% sure tho

Hope this helps =w=

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rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first
adoni [48]

<u>Answer:</u> The products of the given chemical equation are HCO_3^-\text{ and }OH^-

<u>Explanation:</u>

Protonation equation is defined as the equation in which protons get added in the substance.

The chemical equation for the protonation of carbonate ion in the presence of water follows:

CO_3^{2-}+H_2O\rightarrow HCO_3^-+OH^-

By Stoichiometry of the reaction:

1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion

Hence, the products of the given chemical equation are HCO_3^-\text{ and }OH^-

3 0
4 years ago
Number of grams of hcl that can react with 0.500 grams of Al(OH)3
vlabodo [156]

Answer:

0.7g of HCl

Explanation:

First, let us write a balanced equation for the reaction between HCl and Al(OH)3.

This is illustrated below:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:

Molar Mass of Al(OH)3 = 27 + 3(16+1)

= 27 + 3(17) = 27 + 51 = 78g/mol.

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g

Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

From the equation above,

78g of Al(OH)3 reacted with 109.5g of HCl.

Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl

7 0
3 years ago
How would I find the number of protons, neutrons and electrons when I only have the atomic number and atom (assignment says assu
coldgirl [10]

Explanation:

Atomic number , protons and electrons have the same value / their value is same .

But for the neutron there is no specific technique. You have to remember the neutrons of every element

8 0
3 years ago
What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?
natita [175]

This question provides us –

  • Weight of \bf  KCl is = 47 g
  • Volume, V = 375 mL

__________________________________________

  • Molar Mass of \bf   KCl –

\qquad \twoheadrightarrow\bf  39.0983 \times 35.453

\qquad \twoheadrightarrow\bf 74.5513

<u>Using formula</u> –

\qquad \purple{\twoheadrightarrow\bf Molarity _{(Solution)} =  \dfrac{ W\times 1000}{MV}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{ 47 \times 1000}{74.5513\times 375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{47000}{27956.7375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \cancel{\dfrac{47000}{27956.7375}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = 1.68117M

\qquad \pink{\twoheadrightarrow\bf Molarity _{(Solution)}  = 1.7M}

  • Henceforth, Molarity of the solution is = 1.7M

___________________________________________

6 0
2 years ago
WILL MARK BRANILEST
ahrayia [7]

Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

4 0
3 years ago
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