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4 years ago

1. What number on the periodic table determines the number of electrons lost or gained?

Chemistry

1 answer:

vekshin14 years ago

8 0

Answer:

1. atomic #

2.The sugar-water is a homogeneous mixture

Explanation:

1. However, if it has positive ion, then this electron number will go down (ie +2 charge means two electrons have been lost, so the electron/atomic number will go down by two) and vice versa.

2.Sugar dissolves and is spread throughout the glass of water. The sand sinks to the bottom. The sugar-water is a homogenous mixture while the sand-water is a heterogeneous mixture. Both are mixtures, but only the sugar-water can also be called a solution.

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A. what is the hybridization of the central atom in cse2?

elixir [45]

The orbital hybridization of the central carbon atom in CSe2 is sp.

In chemical bonding, atomic orbitals may be combined to form appropriate hybrid orbitals suitable for bonding. The orbitals that combine during hybridization must be close enough in energy.

In the compound Cse2, carbon is the central atom bonded to two selenium atoms. The carbon atom in CSe2 is sp hybridized.

Learn more about orbital hybridization: brainly.com/question/1869903

4 0

2 years ago

If you exhale 7.25×10 to the 24th power molecules of CO2. How many moles of CO3 do you exhale?

Assoli18 [71]

None since CO3 does not exist.

7 0

3 years ago

A gamma ray photon has an energy of 4.75 x 10-14 joules. What is the frequency of this radiation?

Rama09 [41]

The frequency of the radiation is equal to $7.17 \times 10^{19}$ Hertz.

Given the following data:

Photon energy = $4.75 \times 10^{-14}$ Joules

To find the frequency of this radiation, we would use the Planck-Einstein equation.

Mathematically, the Planck-Einstein relation is given by the formula:

$E = hf$

Where:

h is Planck constant.

f is photon frequency.

Substituting the given parameters into the formula, we have;

$4.75 \times 10^{-14} = 6.626 \times 10^{-34} \times F\\\\F = \frac{4.75 \times 10^{-14}}{6.626 \times 10^{-34}}$

Frequency, F = $7.17 \times 10^{19}$ Hertz

4 0

3 years ago

What is the capillary rise of ethanol in a glass tube with a 0.1 mm radius if the surface tension of ethanol is 0.032Jm2 and the

sweet [91]

Answer: The capillary rise(h) in the glass tube is = 0.009m

Explanation:

Using the equation

h = 2Tcosθ/rpg

Given

Contact angle, θ = Zero

h = height of the glass tube=?

T = surface tension = $0.032J/m^2$

r = radius of the tube = 0.1mm =0.0001m

p= density of ethanol = $0.71g/cm^3$

g= $9.8m/s^2$

h = $(2 * 0.032 * cos 0)/( 710*9.8*0.0001)$

h= 0.09m

Therefore the capillary rise in the tube is 0.09m

5 0

3 years ago

PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.

Oduvanchick [21]

i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

$\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).$

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ii. Given the dissolution of some substance

$xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)}$ ,

the Ksp, or the solubility product constant, of the preceding equation takes the general form

$K_{sp} = [B]^y [C]^z$ .

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

$K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}$ .

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iii. Presumably, what we're being asked for here is the molar solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

$1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.$

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.

8 0

3 years ago