When combining with nonmetallic atoms, metallic atoms generally will?

What are the factors affecting qualitative analysis in chemistry

Kitty [74]

Answer:

Techniques and Tests

Qualitative analysis typically measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation.

5 0

2 years ago

Find the mass of 175.4mL of benzene if the density is 0.8786g/ml

bulgar [2K]

Hey there!:

Volume = 175.4 mL

density = 0.8786 g/mL

mass = ?

Therefore:

D = m / V

0.8786 = m / 175.4

m = 0.8786 * 175.4

m = 154.10644 g

hope this helps!

5 0

3 years ago

Read the following chemical equation

Naddika [18.5K]

Answer:

For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3

The mole ration is 4:2 (option 1)

Explanation:

Step 1: The unbalanced equation

Fe + O2 → Fe2O3

Step 2: Balancing the equation

Fe + O2 → Fe2O3

On the left side we have 2x O (in O2) and on the right side we have 3x O (in Fe2O3) . To balance the amount of O on both sides, we have to multiply O2 by 3 and Fe2O3 by 2.

Fe + 3O2 → 2Fe2O3

On the left side we have 1x Fe, On the right side we have 4x (in 2Fe2O3). To balance the amount of Fe we have to multiply Fe (on the left side) by 4.

Now the equation is balanced.

4Fe + 3O2 → 2Fe2O3

For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3

The mole ration is 4:2 (option 1)

8 0

3 years ago

1. Fluorine-21 has a half-life of approximately 5 seconds. How much of a 200 gram sample would remain after 1 minute.

SSSSS [86.1K]

Answer:

Fluorine-21 has a half-life of approximately 5 seconds. How much of a 200 gram sample would remain after 1 minute.

1/4096

4 0

3 years ago

Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to

Natalija [7]

Answer:

1- 3 Moles of CO2

2- 132 g of CO2

3- 105,6 g of CO2

4- Limiting Reagent O2

Products form based on limiting reagent (384g O2) :

CO2: 316,8 g

H2O: 172,8 g

Products form based on C3H8 (132,33 g):

CO2: 396,99 g

H2O: 216,54 g

Explanation:

Atomic Masses:

C: 12

H: 1

O: 16

Molecular weights:

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g O2 ⇒ 132 g CO2

128 g O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For 44 g of CH3H8 ⇒160 g of O2

With 132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

Calculations of the products formed in base of quantity of O2 (limiting reagent):

160 g O2 ⇒ 132 g CO2

384 g O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g O2 ⇒ 72 g H2O

384 g O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

Calculations of the products formed in base of quantity of C3H8 (excess reagent):

44 g C3H8 ⇒ 132 g CO2

132,33 g C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

44 g C3H8 ⇒ 72 g H2O

132,33 g C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

3 0

3 years ago