Chemistry, 15.07.2019 16:30 jocelynmarquillo1

Acube with edge = 1.00” of an unknown pure substance has a mas off 10.0 oz. what is the density of this substance in g/ml? in kg/l?

Answers: 1

Show answers

Answers

Answer from: ethanmel21

Answer 1) The density of the cube will be 17.30 g/mL

We know the edge that is 1 inch, we have to convert it into cc which will be;

1" = 16.38 cc and cc = mL

The given mass is in oz, we need to convert this as well in g;

10 oz = 283.495 g;

We have the formula as Mass = Density / Volume

Substituting the values; Density = (283.495 / 16.38) = 17.30 g/mL

So the density will be 17.30 g/mL

Answer 2) The density of cube will be 1.738 Kg/L

The same procedure of conversion would be done here,

Mass in Kg will be 0.02834 Kg and Volume will be 0.0163 L

On solving with the equation as Mass = Density / Volume;

Density = (0.02834 / 0.0163) = 1.738 Kg/L

Hence, the density will be 1.738 Kg/L

Answer from: Quest

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Another question on Chemistry

Chemistry, 21.06.2019 22:30

Embryos of different species look very similar, which shows that the organisms share a ancestor.

Answers: 1

Answer

Chemistry, 22.06.2019 17:00

Astable electron arrangement for an atom is one that does not easily change. how is this arrangement arrived at? a. valence electrons are transferred or shared to create a full outer shell of electrons. b. valence electrons are discarded into space to create a full outer shell of electrons. c. protons (positive charge) pair with valence electrons (negative charge) to create a strong bond. d. outer shells with valence electrons are transferred or shared.

Answers: 2

Answer

Chemistry, 22.06.2019 18:00

How many moles of oxygen gas are produced from the decomposition of six moles of potassium chlorate

Answers: 3

Answer

Chemistry, 22.06.2019 22:30

What is the value of the standard enthalpy of formation of an element in its most stable form?

Answers: 3

Answer

You know the right answer?

Acube with edge = 1.00” of an unknown pure substance has a mas off 10.0 oz. what is the density of t...