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3 years ago

If you have 3 dozen pennies how many ml of water would they displace

1 answer:

wel3 years ago

3 0

The mass of the displaced fluid can be expressed in terms of the density and its volume, m = ρV. The fluid displaced has a weight W = mg, where g is acceleration due to gravity. Therefore, the weight of the displaced fluid can be expressed as W = ρVg.

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A student mixes a solution of lead (II) nitrate with a solution of potassium iodide and notices the formation of a yellow solid.

ioda

Answer:

Lead (II) iodide

Explanation:

The reaction of lead (II) nitrate, Pb(NO₃)₂ with KI is:

Pb(NO₃)₂(aq) + 2KI(aq) → KNO₃(aq) + PbI₂(s)

This is a typical double-replacement reaction where anions and cations exchange its couple.

All nitrates are solubles, thus, KNO₃ is not the precipitate.

The only possibility of precipitate is PbI₂,

Lead (II) iodide, a yellow and insoluble solid...

5 0

3 years ago

What elements on the periidic table are metalloids<br><br>

butalik [34]

Answer:

Boron (B)

Silicon (Si)

Germanium (Ge)

Arsenic (As)

Antimony (Sb)

Tellurium (Te)

Polonium (Po)

8 0

3 years ago

Give the following for SO2 and BrF5:

frosja888 [35]

Answer:

The given molecules are SO2 and BrF5.

Explanation:

Consider the molecule SO2:

The central atom is S.

The number of domains on S in this molecule is three.

Domain geometry is trigonal planar.

But there is a lone pair on the central atom.

So, according to VSEPR theory,

the molecular geometry becomes bent or V-shape.

Hybridization on the central atom is

$sp^{2}$ .

Consider the molecule BrF5:

The central atom is Br.

The number of domains on the central atom is six.

Domain geometry is octahedral.

But the central atom has a lone pair of electrons.

So, the molecular geometry becomes square pyramidal.

The hybridization of the central atom is $sp^{3} d^{2}$ .

The shapes of SO2 and BrF5 are shown below:

6 0

3 years ago

Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu

borishaifa [10]

Answer:

$\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}$

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹: 2.7 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 2.7 - x x x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}$

2. Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}$

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

2.7 x 0.0441

$K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}$

6 0

4 years ago

The d-metals iron, copper, and manganese form cations with different oxidation states. For this reason, they are found in many o

choli [55]

Answer:

They have electrons in their 3d- and 4s-orbital for bond formation.

Explanation:

d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.

The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.

If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.

If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2

4 0

3 years ago

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