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Aleksandr-060686 [28]
3 years ago
12

Only an experiment can prove:

Chemistry
1 answer:
JulsSmile [24]3 years ago
8 0
If your hypothesis is right
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You can decrease the concentration of a solution by adding more solute. True or False
kifflom [539]

Answer:

True.

Explanation:

Now i'm not 100% sure if this is the answer but i believe depending on what you are making/mixing would depend on the answer, this answer probably won't be the same in context of slim making but it could mean true for other things, so without knowing much i would guess True.

7 0
4 years ago
Read 2 more answers
Now consider the reaction when 45.0 g NaOH have been added. What amount of NaOH is this, and what amount of FeCl3 can be consume
gayaneshka [121]

The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.

The mass of NaOH is 45 g

The molar mass of NaOH = 40 g/mol

The moles of NaOH = mass / molar mass

= 45 / 40

= 1.125

Thus, 1.125 mol NaOH is available

3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl

3 mol of NaOH react with 1 mol of FeCl₃

1.125 moles of NaOH will react with x moles of FeCl₃

x = 1.125 / 3

x = 0.375 mol

0.375 mol FeCl₃ can take part in reaction

The molar mass of FeCl₃ is 162 g/mol

The mass of FeCl₃ = moles × mass

= 0.375 × 162

= 60.75 g

Thus, the amount of FeCl₃, which can be consumed is 60.75 g

6 0
4 years ago
20. The mass of one mole of lead (Pb) atoms is 207.2 g. Use a proportion to calculate the n
AnnZ [28]

Answer:

a) atoms Pb = 4.3595 E22 amu

b) g = ( 6.022 E23 amu ) × ( Mw )

Explanation:

  • m Pb = 207.2 g / mol Pb

sample:

∴ m = 15.00 g

a) atoms Pb = ?

  • 1 mol ≡ 6.022 E23 amu

⇒ atoms Pb = (15.00g Pb)(mol Pb/207.2 g)(6.022 E23 amu/mol)

⇒ atoms Pb = 4.3595 E22 amu

b) relationship:

∴ 1 mol (n) ≡ 6.022 E23 amu......(1)

∴ mass (g) ≡ (Mw)×(n)

⇒ n = g / Mw..........(2)

∴ Mw: molecular weight

(1) = (2):

⇒ g / Mw ≡ 6.022 E23 amu

⇒ g ≡ ( 6.022 E23 amu ) × ( Mw )

8 0
3 years ago
1) De los siguientes cambios indicar cuál es químico y cual es físico:
katrin [286]
  1. fisco
  2. quimico
  3. quimico
  4. quimico
  5. quimico
  6. quimico
  7. fisico
  8. fisico
  9. fisico
  10. quimico

7 0
3 years ago
Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 12.0 ft×15.0 ft×8.60ft.
Damm [24]

<u>Given:</u>

Dimensions of the room= 12 ft * 15 ft * 8.60 ft

<u>To determine:</u>

The amount of HCN that gives the lethal dose in the room with the given dimensions

<u>Explanation:</u>

As per the World Health Organization, the lethal dose of HCN is around 300 ppm

300 ppm = 300 mg of HCN/ kg of inhaled air

Volume of air = volume of room = 12 * 15 *8.6 = 1548 ft³

Now,  1 ft³ = 28316.8 cm³

Therefore, the calculated volume of air corresponds to:

1548 * 28316.8 = 4.383 * 10⁷ cm3

Density of air (at room temperature 25 C) = 0.00118 g/cm3

Thus mass of air corresponding to the calculated volume is

Mass = Density * volume = 0.00118 g/cm3 * 4.383 * 10⁷ cm3

= 5.172*10⁴ g = 51.72 kg

Lethal amount of HCN corresponding to 51.72 kg of air would be.

= 51.72 kg air* 300 mg of HCN/1 kg air = 15516 mg

Ans: Lethal dose of HCN = 15.5 g

5 0
3 years ago
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