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3 years ago

Help out please?????

Chemistry

2 answers:

strojnjashka [21]3 years ago

7 0

A Non-Metal obviously

Alexxandr [17]3 years ago

5 0

Answer:

A non- metal

Explanation:

Hope u got ur answer

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If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the

Sedbober [7]

Answer:

Let me give it a try.

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

=0.005775moles of Ca(OH)2.

From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

=0.00385moles of H3PO4.

Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

=0.072Mol/L.

If this is wrong

then Simply Try The formula for Mixing of solutions

C1V1 = C2V2

0.15 x 38.5 = C2 x (15+38.5)

C2 = 0.11M/L.

7 0

2 years ago

Which general formula represents the functional group in a carboxylic acid?

asambeis [7]

Answer:

The answer is RCOOH

7 0

2 years ago

Read 2 more answers

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

5 0

1 year ago

Which solution is the most concentrated?

katrin [286]

Answer:

2.0 mL of 10.5 M H2O2, where H2O2 has a molar mass of 34 g/mol.

Explanation:

It is most concentrated because it contains 10.5 M of Hydrogen peroxide.

4 0

3 years ago

Read 2 more answers

Calculate the theoretical yield of alum expected from 0.9875 g of aluminum foil. assume the aluminum is the limiting reactant.

skelet666 [1.2K]

Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.

Explanation: Reaction to form alum from Aluminium is given as:

$2Al(s)+2KOH(aq.)+2H_2O(l)+4H_2SO_4(aq.)\rightarrow 2KAl(SO_4)_2(s).12H_2O(l)+3H_2(g)$

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.

By stoichiometry,

2 moles of Al is producing 2 moles of Alum

Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol

Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol

54 g/mol of aluminium will produce 948 g/mol of alum, so

$\text{0.9875 grams of aluminium will produce}=\frac{948g/mol}{54g/mol}\times 0.9875g$

Amount of Alum produced = 17.34 grams

Theoretical yield of alum = 17.34 grams.

5 0

3 years ago