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3 years ago

2 points

Chemistry

1 answer:

Drupady [299]3 years ago

7 0

Answer: The ratio of carbon to bromine atoms in the molecule is 3:1

Explanation:

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass. It can be decomposed into simpler constituents using chemical reactions.

Chemical formula shows the elements in a compound and the relative proportions of those elements.

The chemical formula given for the compound is $C_3H_6BrCl$ which means the ratio of carbon to bromine atoms is 3: 1.

Thus the ratio of carbon to bromine atoms in the molecule is 3:1

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The salt formed by the reaction of the weak acid hydrocyanic acid, HCN, with the strong base potassium hydroxide is

iogann1982 [59]

Answer:

2.28 × 10^-3 mol/L

Explanation:

The equation for the equilibrium is

CN^- + H2O ⇌ HCN + OH^-

Ka = 4.9 × 10^-10

KaKb = Kw

4.9 × 10^-10 Kb = 1.00 × 10^-14

Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5

Now, we can set up an ICE table

CN^- + H2O ⇌ HCN + OH^-

I/(mol/L) 0.255 0 0

C/(mol/L) -x +x +x

E/(mol/L) 0.255 - x x x

Ka = x^2/(0.255 - x) = 2.05 × 10^-5

Check for negligibility

0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6

x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3

[OH^-] = x mol/L = 2.28 × 10^-3 mol/L

5 0

2 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s) + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s) + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s) + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

5 0

3 years ago

A 1.2 kg block of iron at 32 ∘C is rapidly heated by a torch such that 12 kJ is transferred to it. What temperature would the bl

lbvjy [14]

Answer:

For iron

Final temperature = 54,22°C

For copper

Final Temperature = 63.67 °C

Explanation

Hello,

You are using a torch to warm up a block of iron that has an initial temperature of 32°C.

The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.

So you need to use the heat capacity equation as follow in the paper.

The equation has to have all terms in the same units, so:

q = 12000 J

s = 0.450 J / g °C

m = 1200 g

Ti = 32 °C

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mars1129 [50]

Answer:

Explanations:

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1 year ago

ANSWER ASAP AND I WILL GIVE BRAINLYEST!!!!

Vlada [557]

Answer:

20m

Explanation:

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3 years ago