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3 years ago

Give an example of a base in chemistry

Chemistry

2 answers:

slega [8]3 years ago

8 0

An example of a base in chemistry would be like soap or lye

vivado [14]3 years ago

4 0

Acetone which is a weak Lewis base. Acetone: C3H6O

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Because they have aromas, compounds with a ring of resonance bonds are called –

Shalnov [3]

Answer:

the answer is Fragrent compounds

6 0

3 years ago

Be sure to answer all parts. Calculate the molality, molarity, and mole fraction of FeCl3 in a 24.0 mass % aqueous solution (d =

Anna71 [15]

Answer:

m= 1.84 m

M= 1.79 M

mole fraction (X) =

Xsolute= 0.032

Xsolvent = 0.967

Explanation:

1. Find the grams of FeCl3 in the solution: when we have a mass % we assume that there is 100 g of solution so 24% means 24 g of FeCl3 in the solution. The rest 76 g are water.

2. For molality we have the formula m= moles of solute / Kg solvent

so first we pass the grams of FeCl3 to moles of FeCl3:

24 g of FeCl3x(1 mol FeCl3/162.2 g FeCl3) = 0.14 moles FeCl3

If we had 76 g of water we convert it to Kg:

76 g water x(1 Kg of water/1000 g of water) = 0.076 Kg of water

now we divide m = 0.14 moles FeCl3/0.076 Kg of water

m= 1.84 m

3. For molarity we have the formula M= moles of solute /L of solution

the moles we already have 0.14 moles FeCl3

the (L) of solution we need to use the density of the solution to find the volume value. For this purpose we have: 100 g of solution and the density d= 1.280 g/mL

The density formula is d = (m) mass/(V) volume if we clear the unknown value that is the volume we have that (V) volume = m/d

so V = 100 g / 1.280 g/mL = 78.12 mL = 0.078 L

We replace the values in the M formula

M= 0.14 moles of FeCl3/0.078 L

M= 1.79 M

3. Finally the mole fraction (x) has the formula

X(solute) = moles of solute /moles of solution

X(solvent) moles of solvent /moles of solution

X(solute) + X(solvent) = 1

we need to find the moles of the solvent and we add the moles of the solute like this we have the moles of the solution:

76 g of water x(1 mol of water /18 g of water) = 4.2 moles of water

moles of solution = 0.14 moles of FeCl3 + 4.2 moles of water = 4.34 moles of solution

X(solute) = 0.14 moles of FeCl3/4.34 moles of solution = 0.032

1 - X(solute) = 1 - 0.032 = 0.967

6 0

2 years ago

Jerod used a golf club to hit a ball what was the action and reaction

Triss [41]

The action was him hitting the ball the reaction was the ball moving after being hit

6 0

2 years ago

Read 2 more answers

Some radioactive nuclides have very short half-lives, for example, I-31 has a half-life of approximately 8 days. Pu-234, by comp

lorasvet [3.4K]

Answer:

Here's what I find.

Explanation:

Iodine-131

Iodine-131 is both a beta emitter and a gamma emitter.

$_{53}^{131}\text{I}\longrightarrow \, _{54}^{131}\text{Xe} +\, _{-1}^{0}\text{e} +\, _{0}^{0}\gamma$

About 90 % of the energy is β-radiation and 10 % is γ-radiation. Both forms are highly energetic.

The main danger is from ingestion. The iodine concentrates in thyroid gland, where the β-radiation destroys cells up to 2 mm from the tissues that absorbed it.

Both the β- and γ-radiation cause cell mutations that can later become cancerous. Small doses, such as those absorbed from the nuclear disasters in the Ukraine and Japan, can cause cancers years after the original iodine has disappeared.

Plutonium-239

Plutonium-239 is an alpha emitter.

$_{94}^{239}\text{U} \longrightarrow \, _{92}^{235}\text{Xe} + \, _{2}^{4}\text{He}$

Alpha particles cannot penetrate the skin, so external exposure isn't much of a health risk.

However, they are extremely dangerous when they are inhaled and get inside cells. They travel first to the blood or lymph system and later to the bone marrow and liver, where they cause up to 1000 times more chromosomal damage than beta or gamma rays.

It takes about 20 years for plutonium to be eliminated from the liver around 50 years for from the skeleton, so it has a long time to cause damage.

6 0

3 years ago

Does pb(no3)2 + Na3(PO4) = Pb3(PO4)2 + Na(NO3) have a precipitate?

ipn [44]

Answer:

Yes, Pb3(PO4)2.

Explanation:

Hello there!

In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.

Regards!

4 0

3 years ago