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MatroZZZ [7]
2 years ago
8

1. What is the oxidation number for the silver ion in tarnish?

Chemistry
1 answer:
spayn [35]2 years ago
7 0
Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1
You might be interested in
A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells.
Mars2501 [29]

Answer:

The correct answer is "1.0100".

Explanation:

Let the volume of mixture be 100 ml.

then,

The volume of DMSO will be 10 mL as well as that of water will be 90 mL.

DMSO will be:

= 10\times 1.1004

= 11.004 \ g

The total mass of mixture will be:

= 90+11.004

= 101.004 \ g

Density of mixture will be:

= \frac{Mass}{Volume}

= \frac{101.004}{100}

= 1.01004 \ g/mL

hence,

Specific gravity of mixture will be:

= \frac{Density \ of \ mixture}{Density \ of \ water}

= \frac{1.01004}{1}

= 1.0100

3 0
2 years ago
Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the
bulgar [2K]
0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule. 
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
<span>Therefore moles of NaOH = 0.02 </span>
<span>conc = moles / volume </span>
<span>Conc NaOH = 0.02 / 0.025L = 0.8M </span>

4 0
2 years ago
Read 2 more answers
A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0
2 years ago
What mitigation measures can communities do to reduce the damage and impact of sudden geologic hazards?
jasenka [17]

Explanation:

require an emergency support immediately

4 0
2 years ago
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