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levacccp [35]
3 years ago
7

Which of these changes would be classified as producing a chemical change a. freezing water b. mixing salt and water to make a s

olution c. heating cooking oil d. mixing a acid and a base to make a salt
Chemistry
1 answer:
trasher [3.6K]3 years ago
4 0

Answer:

Option B , Option C and Option D

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Weak acids and weak bases react with metals and have the ability to corrode metals.
cestrela7 [59]

Answer:

True

Explanation:

6 0
3 years ago
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How many molecules of water can be produced from 46.72 liters of C8H18? 2\:C_8H_{18}\:+\:25\:O_2\:\:\Longrightarrow\:\:16\:CO_2\
Dmitry [639]

Answer:

1.13×10^25 molecules of water.

Explanation:

Equation of the reaction;

C8H18(g) + 25/2 O2 (g) -------> 8CO2(g) + 9H2O(l)

It is important to first put down the balanced reaction equation. It is not possible to solve any problem on stoichiometric relationship without a balanced reaction equation. Once the equation is obtained, we can now proceed with other steps in the solution of the problem.

From the reaction equation, 1 mole of C8H18 produces 9 moles of water

1 mole of C8H18 occupies 22.4L volume while 1 mole of water contains 6.02×10^23 molecules of water

Hence

22.4 L of C8H18 produces 9(6.02×10^23) molecules of water

46.72 L of C8H18 will produce 46.72 L × 9(6.02×10^23) molecules of water/22.4 L

= 113×10^23 or 1.13×10^25 molecules of water.

5 0
3 years ago
Can somebody answer this question for me it might be easy for you answer part 1,2,and 3
jonny [76]

Answer:

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8 0
3 years ago
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One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) techniq
frozen [14]

Answer:

a. i. 8.447 × 10⁻³ T ii.  27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

a.

i. What strength of magnetic field is required?

Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

F = F'

Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

So,

B = mv/rq

B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)

B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

B = 0.8447 × 10⁻² kg/sC)

B = 8.447 × 10⁻³ T

(ii) What is the diameter of the 13C semicircle?

Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and  = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)

r' = 1.357 × 10⁻¹ kgm/TC)

r' = 0.1357 m

r' = 13.57 cm

Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm

b.

i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?

Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

7 0
2 years ago
Which of the following is unchanged at the end of the CNO cycle?
Musya8 [376]

Answer:

The correct option is: A. carbon-12

Explanation:

The CNO cycle, the abbreviation for the carbon-nitrogen-oxygen cycle, is a catalytic cycle by which the stars produce helium from elemental hydrogen, via a series of nuclear fusion reactions.

This cycle involves the fusion of four protons with carbon (_{6}^{12}\textrm{C}), nitrogen isotope (_{7}^{13}\textrm{N}), and oxygen isotope (_{8}^{15}\textrm{O}), to give an alpha particle and two electron neutrinos and positrons.

The reaction involves the regeneration of carbon (_{6}^{12}\textrm{C}) nucleus in the last step.

_{6}^{12}\textrm{C} → _{7}^{13}\textrm{N} → _{6}^{13}\textrm{C} → _{7}^{14}\textrm{N} → _{8}^{15}\textrm{O} → _{7}^{15}\textrm{N} → _{6}^{12}\textrm{C}

3 0
3 years ago
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