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Ede4ka [16]
3 years ago
15

PLEASE HELP!!! 20 POINTS!!!

Chemistry
2 answers:
Sergio [31]3 years ago
8 0
1. True
2. True
3. False
4. True
8090 [49]3 years ago
5 0
1. False
2.False
3.Fasle
4.True
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Agustina was standing on the edge of a cliff of unknown height holding a bowling ball. She wanted to find the height of the clif
nlexa [21]

Answer:

433 m

Explanation:

Since the fall represents motion under gravity, we use the equation

s = ut - 1/2gt² where s = height of cliff or distance bowling ball falls through, u = initial velocity of bowling ball = 0 m/s(since it starts from rest), t = time = 9.4 s and g = acceleration due to gravity = -9.8 m/s².

So, substituting the values of the variables into the equation, we have

s = 0 m/s × 9.4 s - 1/2 × 9.8 m/s² × (9.4 s)²

s = 0 m - 1/2 × -9.8 m/s² × 88.36 s²

s = 1/2(865.928 m)

s = 432.964

s ≅ 433 m

5 0
3 years ago
Write the full ground state electron configuration of C − C− . electron configuration: Which neutral atom is isoelectronic with
stepan [7]

Answer: 1. Is2 2s2 2p3

2. Nitrogen

Explanation: The number of electron present In C = 6

But an extra electron is added since the charge on C is -1, this therefore makes the total electron 7.

1. By arrangement, the Electronic configuration is therefore;

Ans: 1s2 2s2 2p3

2. It is explained how C has 7 electrons, we can proceed then.

Neutral atom have atomic number of 7.

The element with atomic number of 7 is;

Ans: NITROGEN

7 0
3 years ago
A reaction between ethene and oxygen is used to produce epoxyethane. When silver is combined with ethene, the reaction for produ
taurus [48]
I think the answer is d but I’m not sure
3 0
3 years ago
Read 2 more answers
Calculate the mass in grams of carbon needed to produce 160 g of CO2 in burning: C + O2 --> CO2
Vladimir [108]
160gof CO2×1molof CO2\44g of CO2×1mole of C\1mol of CO2×12g of C\1mole of C=
43.63g of C
5 0
3 years ago
The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) th
seropon [69]

Answer:

(a) a = 5.08x10⁻⁸ cm

(b) r = 179.6 pm  

Explanation:

(a) The lattice parameter "a" can be calculated using the following equation:

\rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}}      

<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>N_{A}<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>

Hence the lattice parameter is:  

a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3}

a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm

(b) We know that the lattice parameter of a FCC structure is:

a = \frac{4r}{\sqrt{2}}

<em>where r: is the atomic radius of Th</em>

Hence, the atomic radius of Th is:

r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm    

I hope it helps you!    

4 0
3 years ago
Read 2 more answers
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