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3 years ago

73.5 grams C, 4.20 grams H, and 72.3 grams Cl. What is the percent composition of each element in this compound?

1 answer:

6 0

To find percent composition do "parts over the whole" ---> each individual mass over the total mass times 100 (since it is a percentage).

total mass ("the whole")= 73.5 + 42.0 + 72.3 grams= 150 grams

1. Carbon %= 73.5 grams/ 150 grams x 100= 49.0%

2. Hydrogen %= 4.20 grams/ 150 grams x 100= 2.80%

3. Chlorine %= 72.3 grams/ 150 grams x 100= 48.2%

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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:

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

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

3 years ago

Put the following names in the correct alphabetic indexing order:(1) Topper & Casey Plumbing(2) KST Enterprises(3) Leland an

dangina [55]

Answer: the correct option is 2, 3, 4, 1.

Explanation:alphabetic indexing order is the order in which files or names are being arranged according to the alphabet. In the following names:

2) KST Enterprises

3)Leland and Son Graphics

4)Lucinda Topper

1) Topper & Casey Plumbing.

While arranging alphabetically, the first letters are usually considered but in a scenario where alphabet occurs twice( 3 And 4) the second letter is considered. I hope this helps, thanks

5 0

3 years ago

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DanielleElmas [232]

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3 years ago

Read 2 more answers

How many atoms of each element are in the chemical formula P2O5?

marshall27 [118]

Answer:A

Explanation:2 phosphorus and 2 oxygen

4 0

3 years ago

A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B = 21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A: M = m1 *0.8662 O = m1 *0.1338

compound B: M = m2 *0.90666 O = m2 *0.09334

compound C: M = m3 *0.92832 O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338 m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48 ⇒ m2 = 227.74 g

0.92832m3 = 206.48 ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0

3 years ago