What is The compound with the formula SiCl4

Give two examples of energy transfer by conduction between two<br> objects

algol13

A boat being accelerated by the engine

Bringing water to the boil in an electric kettle

7 0

3 years ago

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

If a players natural resistance to an unbalanced force is called

jonny [76]

Answer:

Explanation:

i don't know if this is 100% correct but i think it's inertia

7 0

2 years ago

26 POINTS AND BRAINLY

Mandarinka [93]

C. The answer is c.

4 0

3 years ago

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A student balances the following redox reaction using half-reactions.

ICE Princess25 [194]

Answer: 6.

Explanation:

1) Aluminum

$Al^0-3e^----\ \textgreater \ Al^{3+}$

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.

2) Manganesium

$Mn^{2+}+2e^{-}---\ \textgreater \ Mn$

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

$2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}

3Mn^{2+}+6e^{-}---\ \textgreater \ 3Mn^{0}$

4) Net equation

Add the two half-equations:

$2Al^{0}+3Mn^{2+}----\ \textgreater \ 2Al^{3+}+3Mn^{0}$

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.

The right side has 2 Al, 3 Mn, and 2*3 positive charges.

So, the equation is balanced.

5) Count the number of electrons involved.

As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.

5 0

2 years ago

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