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andre [41]
3 years ago
13

Can alkenes react with potassium dichromate?

Chemistry
1 answer:
Lady bird [3.3K]3 years ago
3 0
<span>alkenes donot react with potassium dichromate , but can react with potassium permanganate</span>
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Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

r = \dfrac{a}{2\sqrt{2}}

where;

a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

3 0
2 years ago
What is the particle that is labeled with a question mark (?) in the diagram?
Nat2105 [25]

Answer is: quark.

Quark is a type of elementary particle and a fundamental constituent of matter.

Quarks form composite hadrons (protons and neutrons). Protons and neutrons are in the nucleus of an atom.

Hadrons include baryons (protons and neutrons) and mesons.

There are six types of quarks: up, down, strange, charm, bottom, and top.

5 0
3 years ago
Read 2 more answers
For a particular reaction, ΔH=−111.4 kJ/mol and ΔS=−25.0 J/(mol·K).
PtichkaEL [24]
That’s the answer ^^^^
3 0
3 years ago
to find the density of stopper I weighted it and found its mass to 4.8g. After that I filled a graduated cylinder with 32.1mL of
Fynjy0 [20]

Answer:

0.68g/ml

Explanation:

The density of an object is its mass per unit volume. It is calculated using the formula

Density = mass / volume

Mass of stopper weighed = 4.8g

The volume of stopper can be got by subtracting the (volume of water) from the (volume of water+stopper) i.e.

= 39.2ml - 32.1ml

= 7.1ml

Volume of stopper = 7.1ml

Density of stopper= 4.8/7.1

Density= 0.676056

Therefore, the density of the stopper is 0.68g/ml

4 0
3 years ago
A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH so
uranmaximum [27]

Answer:

<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>

Explanation:

[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>

Ionic equation:

{ \bf{2OH { }^{ - }  _{(aq)} + 2H { }^{ + } _{(aq)}→  2H _{2} O _{(l)} }}

first, we get moles of potassium hydroxide in 28.94 ml :

{ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

{ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\  = { \underline{423.95 \: g \approx424 \: grams}}

for the molar mass:

8 0
2 years ago
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