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3 years ago

A log has a density of .8 g/cm³. What will happen to this log in freshwater, which has a density of 1.0 g/cm³?

Chemistry

2 answers:

Lunna [17]3 years ago

8 0

Answer:

B: It will float.

Explanation:

on edge2021! hope this helps!!~ (=^▽^=)

Inga [223]3 years ago

6 0

Answer:

The log will float on the water because its density is lower than the liquid, so it will stay at the top due to Archimedes' principle.

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Among the factors that affect the solubility of a substance are type of solvent and temperature.

nlexa [21]

Yes there is by the way are there any answers to choose from

7 0

3 years ago

What is the mass in grams of ba(io3)2 can be dissolved in 500 ml of water at 25 degrees celcius?

lesya [120]

The mass of Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g

<h3>What mass of Ba(IO3)2 can be dissolved in 500 ml of water at 25 degrees celcius?</h3>

The Ksp of Ba(IO3)2 = 1.57 × 10^-9

Molar mass of Ba(IO3)2 = 487 g/mol?

Dissociation of Ba(IO3)2 produces 3 moles of ions as follows:

$Ba(IO_{3})_{2} \leftrightharpoons Ba^{2+} + 2\:IO_{3}^{-}$

$Ksp = [Ba^{2+}]*[IO_{3}^{-}]^{2}$

$[Ba(IO_{3})_{2}] = \sqrt[3]{ksp} =\sqrt[3]{1.57 \times {10}^{ - 9} } \\ [Ba(IO_{3})_{2}] = 1.16 \times {10}^{-3} moldm^{-3}$

moles of Ba(IO3)2 = 1.16 × 10^-3 × 0.5 = 0.58 × 10^-3 moles

mass of Ba(IO3)2 = 0.58 × 10^-3 moles × 487 = 2.82 g

Therefore, mass Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g.

Learn more about mass and moles at: brainly.com/question/15374113

#SPJ12

7 0

2 years ago

HELP PLEASE THIS IS CONFUSING

Rasek [7]

Number of moles = 8.3 x 10⁻¹³

<h3>Further explanation </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

5 x 10¹¹ atoms of Silver :

$\tt mol=\dfrac{5\times 10^{11}}{6.02\times 10^{23}}=8.3\times 10^{-13}$

5 0

3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

3 years ago

Complete the mechanism for the base-catalyzed opening of the epoxide below by adding any missing atoms, bonds, charges, nonbondi

worty [1.4K]

Complete Question:

check the first image for complete part of the question

Answer and Explanation:

Epoxide is a three membered ring made up of two carbon atoms and one oxygen atom. Epoxides are cyclic ethers. Due to its ring size, it is highly strained and very reactive. Epoxide ring opening takes place with respect to addition of acid and base.

Ring opening of epoxide with acid:

In the presence of base, the nucleophile attacks the epoxide ring at more substituted site and inverse stereochemistry takes place.(check file 2 attached)

Ring opening of epoxide with base:

The backside attack of nucleophile takes place in less substituted site and then it undergoes protonation to form a product.

(check file 2 attached)

3 0

3 years ago