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avanturin [10]
3 years ago
8

A sample of helium gas is heated from 20.0°C to 40.0°C. This

Chemistry
1 answer:
Montano1993 [528]3 years ago
8 0

Answer:

C. 548mL

Explanation:

Attached is an image of the explanation. Hope this helps!

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How many grams are in 4.5 x 10^22 molecules of water
Luden [163]

Answer:

1.35 g

Explanation:

water is h2o, so the molar mass is 1.01x2+16.00=18.02. divide 4.5 x 10^22 by 6.022 x 10^23 to get 7.5 x 10^-2 (2 sig figs). 18.02 x 7.5 x 10^-2 is 1.35 g

8 0
2 years ago
How much water should be added to 4.3 moles of LiBr to prepare a 2.05 m solution?
arsen [322]

Answer:

2.1 kg of water

Explanation:

Step 1: Given data

  • Moles of lithium bromide (solute): 4.3 moles
  • Molality of the solution (m): 2.05 m (2.05 mol/kg)
  • Mass of water (solvent): ?

Step 2: Calculate the mass of water required

Molality is equal to the moles of solute divided by the kilograms of solvent.

m = moles of solute/kilograms of solvent

kilograms of solvent = moles of solute/m

kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg

8 0
2 years ago
Based on the activity series of metal, which reaction with water will not happen?
laila [671]

Answer:

B

Explanation:

B, H2O + Na The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive. Lithium, sodium, and potassium all react with water,

5 0
2 years ago
Determine the specific heat ofmaterial if a 12g sample absorbed 48j as it was heated from 20-40
devlian [24]

Answer:

c =0.2 J/g.°C

Explanation:

Given data:

Specific heat of material = ?

Mass of sample = 12 g

Heat absorbed = 48 J

Initial temperature = 20°C

Final temperature = 40°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  40°C -20°C

ΔT =  20°C

48 J = 12 g×c×20°C

48 J =240 g.°C×c

c = 48 J/240 g.°C

c =0.2 J/g.°C

6 0
2 years ago
What is the density of an object with a mass of 16 g and 3.0 ml
fomenos

Answer:

D = 5.3 g/mL

Explanation:

Density = Mass over Volume

D = m/V

Step 1: Define

D = unknown

m = 16 g

v = 3.0 mL

Step 2: Substitute and Evaluate

D = 16 g / 3.0 mL

D = 5.333333333 g/mL

Step 3: Simplify

We have 2 sig figs.

5.333333333 g/mL ≈ 5.3 g/mL

6 0
3 years ago
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