How many total atoms are in 0.670 g of p2o5?

Select the word or phrase from the drop-down menu to describe ionic compounds. A formula unit represents the simplest ratio of e

VMariaS [2998]

Response:

CRYSTAL

A LARGE NUMBER OF ATOMS ORGANIZED IN A REGULAR STRUCTURE

1:1

Reasoning:

8 0

2 months ago

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A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

KiRa [2933]

The unknown acid is identified as either butanoic acid or ascorbic acid. To ascertain the number of moles based on the given molarity, we utilize the following relationship: Molarity of NaOH solution = 0.570 M and Volume of solution = 39.55 mL. Utilizing the values in the provided equation, we derive the necessary data. The equation governing NaOH and monoprotic acid reactions indicates that one mole of NaOH reacts with one mole of HX, resulting in 0.0225 moles of the monoprotic acid. Conversely, in the case of NaOH and diprotic acid interactions, the stoichiometry is such that two moles of NaOH engage with one mole of diprotic acid. Consequently, we can calculate moles for butanoic acid with a mass of 2.002 g and a molar mass of 88 g/mol, leading us to the conclusion that both butanoic and ascorbic acids represent the unknown acid being neutralized.

5 0

2 months ago

In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2. What i

VMariaS [2998]

Clarification:

Within a hexagonal close-packed (HCP) unit cell, the ratio of lattice points to octahedral and tetrahedral holes is 1: 1: 2.

Defining the:

Number of lattice points = 1x.

Number of octahedral points = 1x

Number of tetrahedral points = 2x

Assuming anions occupy the HCP lattice points and cations fill half of the tetrahedral holes.

The number of anions in the HCP lattice points, A= 1x

The cation count at tetrahedral points, B = $\frac{2x}{2}=x$

Thus, the compound's formula will be = $A_{1x}B_{1x}=AB$

If anions occupy the HCP lattice points and cations take all of the tetrahedral holes, the number of anions remains as A= 1x

The count of cations at tetrahedral points would now be B = 2x

Thus, the formula for the compound will be = $A_{1x}B_{2x}=AB_2$

6 0

1 month ago

What is the volume in ml of a 1.11 carat diamond, given the density of diamond is 3.51 g/ml?

KiRa [2933]

The answer is that the diamond's volume measures 0.063 ml.
With a density of d(diamond) = 3.51 g/ml, and a mass of m(diamond) = 1.11 carat, with 1 carat being equivalent to 0.2 grams, we convert m(diamond) to grams: m(diamond) = 1.11 carat·0.2 gram/carat, which gives m(diamond) = 0.222 g.
To find the volume: V(diamond) = m(diamond) ÷ d(diamond), which results in V(diamond) = 0.222 g ÷ 3.51 g/ml = 0.063 ml.

3 0

2 months ago