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Scorpion4ik [409]
3 years ago
11

Which of the following would be a strong base? HNO3 LiOH NaNO3 KHSO4

Chemistry
1 answer:
lakkis [162]3 years ago
7 0
I would say the last one because it includes more elements
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How would you detect sulphur in organic compounds​
natita [175]

Answer:

sodium fusion is changed to sodium sulphide if sulphur is present in the compound

8 0
3 years ago
How many moles are there in 13g of KMnO4?
Hoochie [10]

Answer:

0.08 moles

Explanation:

Relative atomic mass of KMnO4 = K + Mn + O4 = 39 + 55 + 16 x 4 = 158

=> Moles in 13g of KMnO4 = \frac{Mass}{Relative-atomic-mass} = \frac{13}{158} = 0.082278.. = 0.08 moles

3 0
3 years ago
Read 2 more answers
Thank uuuu so much!!!!!
kow [346]

Answer:

A. 0.83

Explanation:

Hopefully this helps!

7 0
3 years ago
Describe the particle mode of matter. what does the brownian motion tells us about the particles in matter?give other evidence t
svetlana [45]

Answer:

dium (a liquid or a gas). This pattern of motion typically consists of random fluctuations in a particle's position inside a fluid sub-domain, followed by a relocation to another sub-domain. Each relocation is followed by more fluctuations within the new closed volume. This pattern describes a fluid at thermal equilibrium, defined by a given temperature. Within such a fluid, there exists no preferential direction of flow (as in transport phenomena). More specifically, the fluid's overall linear and angular momenta remain null over time. The kinetic energies of the molecular Brownian motions, together with those of molecular rotations and vibrations, sum up to the caloric component of a fluid's internal energy (the Equipartition theorem).

Explanation:

3 0
2 years ago
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
3 years ago
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