All voltmeters have two probes attached to make a measurement explain why you cannot make a voltmeter with a single probe to measure the voltage of a wire

Answers

Answer 1

Answer:

As voltages is a potential in relation to a reference, one probe must be on the reference or "zero" planes and the other must be on the point being measured.

Why does a voltmeter not accurately read voltage?

because the voltmeter uses some of the main circuit's current. Main present in the circuit diminishes as a result, and the voltmeter's reading of the potential difference does not correspond to its true value.

Why are there two probes on a voltmeter?

Nothing is measured at a specific point by the voltmeter. It gauges the voltage (V) differential between two circuit locations. Thus, a multimeter has two leads rather than one.

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What you understand by macrocylic effect in coordination chemistry

Answers

Explanation:

The macrocyclic effect in coordination chemistry refers to the enhanced stability of metal complexes that have a macrocyclic ligand. A macrocyclic ligand is a ligand that forms a ring around the metal ion, forming a macrocyclic complex. This structure increases the enthalpy of the complex, making it more thermodynamically stable. In other words, the macrocyclic effect increases the stability of a metal complex by making the ligand more difficult to remove. This effect is especially important in biological systems, where macrocyclic ligands play a key role in many enzymatic reactions.

According to the valence bond theory the triple bond in ethyne consists of

Answers

Answer:
According to the valence bond theory the triple bond in ethyne consists of one sigma bond and two pi bonds.

Explanation:
Atomic number of carbon is 6. The ground state electronic configuration of carbon is as follow,

1s², 2s², 2p²

And the excited state electronic configuration of carbon is as follow,

1s², 2s¹, 2px¹, 2py¹, 2pz¹

In ethyne the 2s¹ orbital and 2px¹ orbitals having unpaired electrons form sigma bonds by head to head overlapping with orbitals of hydrogen atom and carbon atom. The remaining 2py¹ and 2pz¹ orbitals of both carbons overlap perpendicular to the existing sigma bond resulting in the formation of two pi bonds.

In the United States, volume of irrigation water is usually expressed in acre-feet. One acre-foot is a volume of water sufficient to cover 1 acre of land to a depth of 1 ft (640 acres = 1 mi2, 1 mi = 5280 ft). The principal lake in the California Water Project is Lake Oroville, whose water storage capacity is listed as 3.54 x 106 acre-feet. Express the volume of Lake Oroville in (a) cubic feet, (b) cubic meters, and (c) US gallons.

Answers

Answer:

a) V = 1.542 E11 ft³

b) V = 4367011968 m³

c) V = 1.1535 E12 us gal

Explanation:

Acre surface is defined as 66 by 660 feet at a depth of one foot:

⇒ Vacre-foot = 66ft*660ft*1ft = 43560 ft³

lake:

a) V = 3.54 E6 acre-feet * ( 43560 ft³ / acre-foot ) = 1.542 E11 ft³

b) V = 1.542 E11 ft³ * ( 0.02832m³ / ft³ ) = 4367011968 m³

c) V = 1.542 E11 ft³ * ( 7.48052 us gal/ft³ ) = 1.1535 E12 us gal

The length of the marathon race is approximately 26.2 mi. What is the distance in kilometers?

Answers

Answer : The distance in kilometers is, 42.2 km

Explanation :

As we are given that the length of the marathon race is 26.2 mile. Now we have to determine the distance in kilometers.

The conversion used for distance from mile to kilometer is:

1 mile = 1.609 km

As, 1 mile = 1.609 km

So, 26.2 mile = $\frac{26.2\text{ mile}}{1\text{ mile}}* 1.609km$

= 42.2 km

Thus, the distance in kilometers is, 42.2 km

Answer: 42.16481

Explanation:

What is the coupling constant of the alkene signals of the NMR of your product? (The two alkene signal are doublets at 7.28 and 7.14 ppm.) How does this information tell you that you isolated the trans and not cis product?

Answers

Answer:

Explanation:

I need more information to answer this question and a better figure.

1. What is the frequency of the NMR machine?

Possible solution:

1. coupling constant Jab (in ppm) is given by

\nu_{a} - \nu_{b} = 4J_{ab}

2. Jab (in ppm) * Frequency of machine in (MHz)/106 is Jab in Hz

3. for cis vicinal Hydrogen Jab = 6-14 Hz

4. for trans vicinal Hydrogens Jab = 11-18 Hz

Now, considering 2 doublets are centered at 7.14 and 7.28 ppm , it gives

7.14 -7.28 = 4 Jab

thus, Jab = 0.07 ppm

Now if we consider a 100 MHz machine,

Jab = 7 Hz , thus indicating cis product

but if machine is 300 MHz

then Jab = 21 Hz , thus indicating a trans product.

But, most probably I feel it is a trans product. I hope it helps.

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Answers

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer:

$m_(PbI_2)=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_(Pb(NO_3)_2)=(0.14gPb(NO_3)_2)/(1g\ sln)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2) *(1.134g\ sln)/(1mL\ sln) *96.7mL\ sln\n\nn_(Pb(NO_3)_2)=0.04635molPb(NO_3)_2\n\nn_(KI)=(0.12gKI)/(1g\ sln)*(1molKI)/(166.0gKI) *(1.093g\ sln)/(1mL\ sln) *99.8mL\ sln\n\nn_(KI)=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_(PbI_2)=0.07885molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \n\nm_(PbI_2)=18.2gPbI_2$

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