Which label belongs in the area marked X?

A sample of 508.4 grams of copper completely reacted with oxygen to form 572.4 grams of a copper oxide product. how many grams o

Svet_ta [14]

According to law of conservation of mass, mass can neither be destroyed nor created in a chemical reaction. Thus, sum of masses of reactants must be equal to sum of masses of products in a reaction.

The chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, sum of masses of Cu and oxygen gas should be equal to CuO formed.

$2m_{Cu}+m_{O_{2}}=2m_{CuO}$

Thus, mass of oxygen will be:

$m_{O_{2}}=2(572.4-508.4)g=128 g$

This can be further proved as follows:

The balanced chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, 2 moles of Cu completely reacts with 1 mole of $O_{2}$ to give 2 moles of $CuO$ .

Thus, 1 mole of Cu reacts with 0.5 moles of $O_{2}$ .

The mass of Cu is 508.4 and molar mass is 63.546 g/mol, number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{508.4 g}{63.546 g/mol}=8 mol$

Thus, number of moles of $O_{2}$ reacting will be:

$n_{O_{2}}=8\times 0.5 mol=4 mol$

Molar mass of oxygen molecule is 32 g/mol thus, mass can be calculated as follows:

m=n×M=4 mol×32 g/mol=128 g/mol

This satisfies the law of conservation of mass.

7 0

2 years ago

In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.6 L

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}$

${V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L$

${V_2}=0.27\ L$

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0

2 years ago

A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w

MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

$Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}$

Mass of the water = $1 g/mL\times 165 mL=165 g$

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

$w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%$

The weight/weight % or percent by mass of the solute is 5.41 %.

8 0

2 years ago

Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As

Crank

Answer:

1.8 × 10⁻¹⁶ mol

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

1.2 + 3s 2s

$K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\$

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0

2 years ago

There are two types of nucleic acids, dna and rna. nearly all organisms use dna, not rna, as the central repository for genetic

Leya [2.2K]

Found the choices. Pls see attachment.

The statements that explains this phenomenon are:
1) DNA contains adenine as one of its nitrogenous bases.
2) DNA has a double-stranded structure that ensures an accurate mechanism of duplication.

6 0

2 years ago