Balance in basic solution: O2(g) + Cr³+ (aq) → H₂O2 (1) + Cr₂O7²- (aq)

An unknown solution has a pH of 7. How would you classify this solution?

xenn [34]

Solution with a ph of 7 Is neutral

7 0

3 years ago

What volume of o2 is needed to react fully with 720. Ml of nh3

Galina-37 [17]

Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.

<h3>What is volume?</h3>

Volume is the area occupied by the substance and is the ratio of the mass to the density.

At STP, 1 mole of gas occupies 22.4 L of volume

Given,

Volume of ammonia reacted = 0.720 L

The combustion reaction is shown as,

$\rm 4NH_{3} + 5O_{2} \rightarrow 4NO +6H_{2}O$

From the stoichiometry of the reaction, it can be said that,

$(4 \times 22.4)$ L of ammonia reacts with $(5 \times 22.4)$ L of oxygen gas.

So, 0.720 L of ammonia will react with:

$\dfrac{ (5 \times 22.4)}{(4 \times 22.4)} \times 0.720 = 0.9 \;\rm L$

Therefore, the volume of oxygen required is 900 mL.

Learn more about volume here:

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6 0

2 years ago

The Chemistry Cafe was out of bread. The cook went next door to the bakery and bought a loaf of bread which has 33 slices. Then,

maria [59]

In the question, we are told that there are;

A loaf containing 33 slices
A loaf containing 33 slices A package of cheese containing 15 slices

We also know that he is making a sandwich that has 2 pieces of both cheese and bread.

Hence;

Total number of bread and cheese = 33 + 15.

Each loaf should have two pieces of each bread and the cheeses make a total of four pieces.

Therefore he can make = 33 + 15/4 = 12 sandwiches.

4 0

1 year ago

Read 2 more answers

What is the volume of a solid with a mass of 17.4 g and a density of 0.934 g /cm cubed?

Anvisha [2.4K]

Answer:

18.6 g/cm^3

Explanation:

17.4g/0.934g/cm^3

(divide)

5 0

2 years ago

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

8 0

3 years ago