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Tanya [424]
3 years ago
6

How can you tell from the name the types of bonds present in a hydrocarbon?

Chemistry
2 answers:
Aneli [31]3 years ago
7 0

Answer:

Of bond is formed by sharing of electrons - covalent

If bond is formed by donation and by accepting electrons - ionic bond

Bond formed between metals - metallic bond.

Snezhnost [94]3 years ago
7 0

The name of a hydrocarbon dictates whether there are single C-C bonds, double C=C bonds, or triple C bonds. This is done by the name ending in -ane, -ene, and -yne. Let’s use common compound ‘methane’ as an example. Methane has a C-C bond because it ends in -ane. Methane is an alk-ANE. Methane has a C=C because it ends in -ene and is an alk-ENE. lastly, methyne has a triple C bond because it ends in -yne and is an alk-YNE.

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Which of the following is the main function of the Circulatory System?
kolbaska11 [484]

Answer:

B: The transport oxygen and nutrients around your body and help to get rid of wastes

Explanation:

The circulatory system carries oxygen and nutrients to the body, but it can also help get rid of waste (though that isn't a huge function of it).

8 0
3 years ago
Each step in any energy conversion process will _____.
Shkiper50 [21]

the correct answer is dissapate...but it is

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4 0
2 years ago
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

  • 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

Xg = 9.3 g * 0.1 L / 1 L = 0.93 g

8 0
3 years ago
Which process was responsible for producing gases in the early atmosphere?
andre [41]

Answer: The early atmosphere

Explanation: Its early atmosphere was probably formed from the gases given out by volcanoes. It is believed that there was intense volcanic activity for the first billion years of the Earth's existence. The early atmosphere was probably mostly carbon dioxide, with little or no oxygen.

6 0
3 years ago
Read 2 more answers
g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The h
Setler79 [48]

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

3 0
3 years ago
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