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blsea [12.9K]
3 years ago
10

Can somebody please help me ! :(

Chemistry
2 answers:
avanturin [10]3 years ago
8 0

Answer:

325

Explanation:

Helen [10]3 years ago
5 0

Answer:

325

Explanation:

velocity = 3250 Hz x 0.1 m

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What did early scientists assume that the polar caps had in common?
Vilka [71]

Answer:

They assumed they both had water.

Explanation:

Because they only could look at it through telescopes that were not advanced

3 0
3 years ago
Which observations is the best evidence that unknown liquid X is a composed of only pure substance
Angelina_Jolie [31]

Answer:

A substance that is composed only of atoms having the same atomic number is ...  36 grams of an unknown liquid at its boiling point,.

4 0
3 years ago
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
4 years ago
How many mL of 2M stock solution would I use to prepare 0.500 L of 0.1 M NaCl?
nordsb [41]

Answer:

25 mL

Explanation:

Step 1: Given data

  • Concentration of the concentrated solution (C₁): 2 M
  • Volume of the concentrated solution (V₁): ?
  • Concentration of the diluted solution (C₂): 0.1 M
  • Volume of the diluted solution (V₂): 0.500 L

Step 2: Calculate the volume of the concentrated NaCl solution

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.1 M × 0.500 L / 2 M

V₁ = 0.025 L = 25 mL

4 0
3 years ago
1. Calculate the RFM of Fe(NO3)2
miss Akunina [59]

Answer:

The Relative Formula Mass of Fe(NO₃)₂ is 179.8524 grams

Explanation:

The Relative Formula Mass is the mass of one mole of a compound expressed in grams, obtained by adding together the Relative Atomic Masses, RAM, of the elements which makes the compound

The Relative Formula Mass of a compound is the same as its Relative Molecular Mass

The relative formula mass of Fe(NO₃)₂ is given as follows;

The relative atomic mass of Fe = 55.845 amu

The relative atomic mass of nitrogen, N = 14.0067 amu

The relative atomic mass of oxygen, O = 15.999 amu

Therefore, we have;

The formula mass of Fe(NO₃)₂ = (55.845 + 2×(14.0067 + 3×15.999)) amu = 179.8524 amu

The Relative Formula Mass of Fe(NO₃)₂ = 179.8524 grams.

5 0
3 years ago
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