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3 years ago

What is the region of a longitudinal wave where particles are closest together called?

2 answers:

8 0

Compressions
In a longitudinal wave, particles of the medium vibrate in a direction that is parallel to the direction that the wave travels. Places where particles of the medium crowd closer together are called compressions. Places where particles of the medium spread farther apart are called rarefactions.

Ronch [10]3 years ago

4 0

Answer:

compressions

Explanation:

in a longitudinal wave, particles of the medium vibrate in a direction that is parallel to the direction that the wave travels. places where particles of the medium crowd closer together are called compressions. places where particles of the medium spread farther apart are called rarefactions

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If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

Step 1: Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

Step 2: Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

Step 3: Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Calculate Q

Q = [Ca2+] [SO42-]

[Ca2+]= 0.050 M [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

Step 7: Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

Step 8: Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0

3 years ago

Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi

uysha [10]

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is $2.9^0C$

Explanation:

Elevation in boiling point:

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of pure carbon disulfide= $46.2^oC$

$k_b$ = boiling point constant = $2.34^0Ckg/mol$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 35.0 g

$w_1$ = mass of solvent (carbon disulphide) = 397.7 g

$M_2$ = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

$(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}$

$T_b=49.1^0C$

Therefore, the change in boiling point is $(49.1-46.2)^oC=2.9^0C$

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3 years ago

Write a balanced formation equation at standard conditions for the compound HI.

allsm [11]

H₂(g) + I₂(s) = 2HI(g)

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lara31 [8.8K]

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2 years ago