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3 years ago

12

The diagram shows a carrier wave before and after modulation. The diagram best represents which type of modulation? FM phase AM

pulse

2 answers:

defon3 years ago

8 0

the answer is A..hope this helps :))

bixtya [17]3 years ago

5 0

Answer: On Edge it's A. FM

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Gerald's science teacher mixed liquid X and liquid Y, both at room-temperature, in a large beaker. two beakers of clear solution

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Answer:

the answer is C

Explanation:

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3 years ago

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How many acetyl coa molecules may be obtained from oxidation of an 18-carbon fatty acid?

saveliy_v [14]

The answer would be 9

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3 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)

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3 years ago

Which of the following options would be the best for dissolving PbBr2(s)?

Nezavi [6.7K]

Answer:

2) Add a solution of NaBr

Explanation:

Lead (II) bromide is an inorganic powdery substance that has a solubility in water of 0.973 g/100 mL at 20°C. It is insoluble in alcohol but is soluble in alkali, ammonia, NaBr, and KBr

PbBr₂ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)₂ and NaBr

Therefore, the best solution for dissolving PbBr₂(s) is NaBr

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3 years ago

2 SO2(g) + O2(g) + 2 H2O(ℓ) −→ 2 H2SO4(ℓ)

DIA [1.3K]

Answer:

6116g

Explanation:

2SO2(g) + O2(g) + 2H2O(ℓ) −→ 2H2SO4(ℓ)

We want to find the mass in grams of SO2 that is needed to react with 1527 g of O2. First we must convert the grams of O2 to moles of O2 then to moles of SO2 and then to grams of SO2

So first lets find the molar mass of O2

The mass of oxygen according to a periodic table is 15.999

Using this the mass of O2 would be 15.999(2) = 31.988g

Next we need to identify the mole ratio of O2 to SO2

Looking at the equation for 1 mole of O2 there are two moles of SO2

Next we need to find the molar mass of SO2

Again the mass of oxygen is 15.999g and the mass of Sulfur is 32.066

So the mass of SO2 would be 15.999(2) + 32.066 = 64.064g

Now that we have found all the needed conversions :

1 mol O2 = 31.988g

1 mol O2 = 2 mol SO2

1 mol SO2 = 64.064g

We can now use dimensional analysis to calculate the answer.

Kindly check the attached image to see the table. ( sorry if its a bit blurry )
Explanation : The conversions are used to cancel out the units to get to the final unit which is gSO2.

Once the units are cancelled out except for the gSO2 we mutliply and divide based off of what the table says to do.
Here first we divide 1527 by 31.988. We than multiply by 2. Finally we multiply by 64.064 to get the final answer which is 6116gSO2

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2 years ago