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jeka94
2 years ago
6

How many bonds does the alpha carbon form?

Chemistry
1 answer:
vaieri [72.5K]2 years ago
4 0
The carbon molecule usually forms 3 bonds with one lone electron on each atom
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The diagram shows the scales used for recording temperatures. The labels for the scales are missing. 3 thermometers are oriented
QveST [7]

Answer d

Explanation:

8 0
1 year ago
Read 2 more answers
Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio
Paha777 [63]
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
 The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) = 
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

6 0
2 years ago
Choose the option from the drop-down menus that is best described by each statement. Involves combining of nuclei: Produces dang
earnstyle [38]

Answer:

1: Nuclear Fusion

2: Nuclear Fission

3: Nuclear Fusion

4: Both

5: Nuclear Fusion

6: Nuclear Fission

Explanation:

Trust me dawg

8 0
2 years ago
Read 2 more answers
The beaker shown below contains a 0.58 M solution of dye in water. How many moles of dye are there in the beaker?
seropon [69]
According to the equation of molarity:

Molarity= no.of moles / volume per liter of Solution

when we have the molarity=0.58 M and the beaker at 150mL so V (per liter) = 150mL/1000 = 0.150 L

by substitution:
∴ No.of moles = Molarity * Volume of solution (per liter)
                        = 0.58 * 0.150 = 0.087 Moles
7 0
1 year ago
Read 2 more answers
Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are
kolbaska11 [484]

Explanation:

Molarity of copper sulfate solution = 1.00 M

Volume of the copper sulfate solution  = 50.0 mL = 0.050 L

Moles of copper sulfate = n

1.00M=\frac{n}{0.050 L}

n = 0.050 L × 1.00 M= 0.050 mol

1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :

1\times 0.050 mol=0.050 mol of copper

a) Mass of 0.050 moles of copper = 0.050 mol × 63.5 g/mol =3.175 g

b) The identity of the compound which formed after the reaction is copper hydroxide.

c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)

d) CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)..[1]

KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)..[2]

Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)

Common ion both sides are removed. The net ionic equation is given as:

Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)

e) Volume of solution after mixing of both solution,V= 50 mL + 50ml = 100 mL

Mass of final solution ,m= 1 mL

Density of solution ,d= 1 g/mL (same as pure water)

m=d\time V=1 g/ml\times 100 mL = 100 g

Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)

Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

Q=mc\Delta T

Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ

Enthalpy of the reaction = ΔH = \frac{Q}{\text{Moles of copper}}

ΔH = \frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

7 0
1 year ago
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