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Hunter-Best [27]
3 years ago
10

Can someone tell me if I put this in the correct pairs???? Help please!!??

Chemistry
1 answer:
e-lub [12.9K]3 years ago
8 0
Yes i think so



hope i helped
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A 14.3-cm3 sample of tin has a mass of 0.104 kg.
VladimirAG [237]
This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³
7 0
3 years ago
I NEED THIS ASAP Someone please help me
IceJOKER [234]

Answer:

B. Four moles of water were produced from this reaction.

Explanation:

I took the test and got it correct

8 0
3 years ago
Read 2 more answers
Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

  • P₁ = 1.3 kPa
  • P₂ = 5.3 kPa
  • T₁ = 85.8°C = 358.96 K
  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

  • ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
  • ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
  • ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

3 0
3 years ago
How many moles of dipyrithione contain 8.2 x 10^24 atoms of N
Pani-rosa [81]
Dipyrithione is a chemical with formula, C₁₀H₈N₂O₂S₂. This means that each molecule of the substance has two (2) atoms of nitrogen. By using the dimensional analysis and Avogadro's number, equal to 6.022 x 10²³, we calculate for the answer as shown below.

    n = (8.2 x 10²⁴ atoms N)(1 molecule dipyrithione/ 2 atoms of N)(1 mole dipyrithione/ 6.022 x 10²³ molecules dipyrithione)

Simplifying,
  n = 6.8 moles dipyrithione

<em>ANSWER: 6.8 moles</em>
4 0
3 years ago
The density of benzene is 0.879g/mL. The volume (with 1 decimal) of 131.9g sample of benzene is how many mL?
timurjin [86]

Answer:

150.1 mL

Explanation:

Step 1: Given data

  • Density of benzene (ρ): 0.879 g/mL
  • Mass of the sample of benzene (m): 131.9 g
  • Volume of the sample of benzene (V): ?

Step 2: Calculate the volume of the sample of benzene

Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.

ρ = m/V

V = m/ρ

V = 131.9 g/(0.879 g/mL)

V = 150.1 mL

5 0
3 years ago
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