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sp2606 [1]
3 years ago
9

20 pts!!!!!!!!!!

Chemistry
2 answers:
Bess [88]3 years ago
3 0
Increase the pressure that's what im going with
Irina18 [472]3 years ago
3 0

It increases the surface area. Lots of tiny bits have more surface area than the big clump. Like sugar and water. Its easier for tiny clumps to dissolve than a big one

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What would be the effect on the nucleus of an atom if it emitted one alpha particle and one gamma ray
Oksana_A [137]
Unstable, radioactive atom emits alpha and gamma ray to transform into a stable atom.

Emitting an alpha particle, decreases the mass number by four and atomic number by two.
Emitting gamma ray reduces the excess energy of the radioactive atom.
4 0
3 years ago
Which agents cause both chemical and physical weathering?
jeka94

Answer:

iron

Explanation:

4 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
1. The Lewis structure for CO2 has a central ________ atom attached to __________ atoms.
maksim [4K]
Central "carbon" atom

2 oxygen atoms

held together by "covalent" bonds

has a "1s2 2s2 2px1 2py1 2pz0" electron
geometry

carbon atom is "sp" hybridized

8 0
3 years ago
CH4(g) + H2O(g) CO(g) + 3H2(g)
boyakko [2]
Presuming the arrow is between H20 and CO

On the left there are 2 gas moles.
On the right there are 4 gas moles.

The equilibrium will shift to the side with the most no. He gas moles when pressure is decreased.

Therefore the answer is A, since 4>2.

If you have any questions, feel free to ask
5 0
3 years ago
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