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3 years ago

11. A 100% silver ring would be an example of a

Chemistry

2 answers:

Tomtit [17]3 years ago

5 0

Answer:

a pure substance or a compound

chubhunter [2.5K]3 years ago

3 0

Answer:

a pure substance

Explanation:

Any substance made from a single material is pure. Silver is an element just like all the others listed on the periodic table, so a block of silver is pure.

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How many grams of gas must be released from a 32.0 L sample of CO2(g) at STP to reduce the volume to 16.6 L at STP?

Hoochie [10]

Answer:

30.3 g

Explanation:

At STP, 1 mol of any gas will occupy 22.4 L.

With the information above in mind, we calculate how many moles are there in 32.0 L:

32.0 L ÷ 22.4 L/mol = 1.43 mol

Then we calculate how many moles would there be in 16.6 L:

16.6 L ÷ 22.4 L/mol = 0.741 mol

The difference in moles is:

1.43 mol - 0.741 mol = 0.689 mol

Finally we convert 0.689 moles of CO₂ into grams, using its molar mass:

0.689 mol * 44 g/mol = 30.3 g

6 0

3 years ago

A chemist titrates 150.0 mL of a 0.2653 M carbonic acid (H2CO3) solution with 0.2196 M NaOH solution at 25 °C. Calculate the pH

xxTIMURxx [149]

Answer:

9.3

Explanation:

This is long and complicated so get ready

We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)

Let the conjugate base be A- and Carbonic acid be HA

A- + H20 ⇄ HA + OH-

To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.

to get moles of acid we take the concentration and multiply by liters to cancel

.2653 x .150 = .039795 mol HA

Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters

We plug it into the equation and found: .181 L

Now use moles and combined volums to fins concentrarion which is .120 M

Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-

Ka is (10^-3.60) = 2.4E-4

Kb x Ka is 10^-14

Kb = 3.98E-11

Now we know Kb = [HA] [OH] / [A-]

Solve for this through algebra by using x for the values you dont know

youll find x^2 = 3.3E-10

X = 1.8 E -5

this is the OH- concentration

-log [oh] = pOH

pOH = 4.73

We know 14-pOH = ph so pH= 9.3

6 0

3 years ago

What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 40.0 minutes? Assume first-order

yaroslaw [1]

Answer:

See below

Explanation:

.75 = 1/2^(40/h)

log .75 / ( log 1/2) = 40 / h

h = half life = 96.37683 min

8 0

2 years ago

What property of hydrocarbon molecules correlates to the viscosity of a particular fraction when it is cooled to room temperatur

lora16 [44]

Because of differences in molecular structure, the empirical formula remains different between hydrocarbons; in linear, or "straight-run" alkanes, alkenes and alkynes, the amount of bonded hydrogen lessens in alkenes and alkynes due to the "self-bonding" or catenation of carbon preventing entire saturation of the hydrocarbon by the formation of double or triple bonds.

This inherent ability of hydrocarbons to bond to themselves is referred to as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane, and in rarer cases, arenes such as benzene. This ability comes from the fact that bond character between carbon atoms is entirely non-polar, in that the distribution of electrons between the two elements is somewhat even due to the same electronegativity values of the elements (~0.30), and does not result in the formation of an electrophile.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0

3 years ago

Read 2 more answers

Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol

max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;