Write and balance the equation for the neutralization reaction between phosphoric acid and

Eddi Din [679]

Answer:

С. form chains or rings by bonding to itself and other atoms.

Explanation:

Carbon is an element in group 4 of the periodic table with unique bonding properties. Carbon posseses 4 valence electrons in its outer shell. This enables carbon to form covalent bonds with the atoms of other elements e.g. nitrogen, phosphorus, oxygen, hydrogen etc.

Carbon can also combine covalently with other carbon atoms i.e. C-C to form long chains and rings in a process called CATENATION. This unique property of carbon makes it the only element that can form so many different compounds.

5 0

3 years ago

You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q

Flauer [41]

Answer:

6,78 mL of 12,0 wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = $10^{-7,20}$ , thus,

Thus, you need to add:

[H⁺] = $10^{-7,2} -10^{-8,0}$ = 5,31x10⁻⁸ M

The total volume of the pool is:

9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,792x10⁻² moles of H⁺ × $\frac{1H_{2}SO_4 mol}{2H^{+} mol}$ = 8,96x10⁻³ moles of H₂SO₄

These moles comes from:

8,96x10⁻³ moles of H₂SO₄ × $\frac{98,1 g}{1 mol}$ × $\frac{100 gSolution}{12 gH_{2}SO_4 }$ × $\frac{1 mL}{1,080 g}$ =

6,78 mL of 12,0wt% H₂SO₄

I hope it helps!

8 0

4 years ago

How many atoms are in 1.6g C? Answer in units of atoms. (no answer choices were given)

maksim [4K]

Answer:

6.022 ×10(index 23) / 7.5 = 0.8293 ×10(index 23)

Explanation:

molar mass of C = 12gmol

therefore in 12g of C there is one mole or an amount of 6.022 ×10(index 23)

∴12g/6.02210(index 23) ×1.6g

3 0

3 years ago

In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu

Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol = 0.810 g/mL

Volume of 3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol = mass/molar mass = 7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

7 0

3 years ago

How many moles of MgO are produced when .250 mol of Mg reacts completely with O2

nignag [31]

Answer:

0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂

Explanation:

In first place, the balanced reaction between Mg and O₂ is:

2 Mg + O₂ ⇒ 2 MgO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

Mg: 2 moles
O₂: 1 mole
MgO: 2 moles

Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?

$moles of MgO=\frac{0.250 moles of Mg*2 moles of MgO}{2 moles of Mg}$

moles of MgO= 0.250

0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂

3 0

3 years ago