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


Answer 1

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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Find the weighted average of these values. Col1 Value 5.00 6.00 7.00Col2 Weight 75.0 % 15.0 % 10.0 %


The weighted average (Avg) for these values has been 5.35.

The weighted average has been an arithmetic calculation of the mean value for the percent abundance of each value.

Computation for weighted average

The weighted average (Avg) for the values has been given by:


The values have been given,


The weighted average has been given as:


For the given set of values, the weighted average (Avg) has been given as:


The weighted average (Avg) for these values has been 5.35.

Learn more about weighted average, here:




Value   5.00        6.00       7.00

Weight 75.0%     15.0%      10.0 %

We can determine the weighted average of these values using the following expression.

Weighted average = ∑ wi × xi


w: relative weight

x: value

Weighted average = 5.00 × 0.750 + 6.00 × 0.150 + 7.00 × 0.100

Weighted average = 5.35

1. How do you determine how many protons are in a neutral element?



The number of protons in an atom is always equal to the atomic number.


Predict the product when 2-methylbutanol is oxidised with pyridium chlorochromate in dichloromethane (PCC/CH2C12) a. CH3CH2CH(CH3)COOH b. CH3CH2CH2CH(CH3)CHO C. CH3CH2CH(CH3)CHO d. CH3CH2CH(CH3)OH 8:40 PM Type a message​





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In this case, according to the process for the one-step oxidation of a primary alcohol with a moderately strong oxidizing agent like pyridinium chlorochromate (PCC), whereby an aldehyde is produced, we infer that the corresponding product will be 2-methylbutanal, which matches with the choice c. CH3CH2CH(CH3)CHO according to the following reaction:

CH_3CH_2CH(CH_3)CH_2OH\rightarrow CH_3CH_2CH(CH_3)COH


What is the difference between a volume that is delivered and a volume that is contained?



When the graduation line denotes the volume contained in the calibrated vessel, the ware is marked “TC”. When the graduation line indicates the volume delivered from the vessel, the ware is marked “TD”.

The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (M) Initial Rate (10–3 M/s) 0.4 0.4 0.2 160 0.2 0.4 0.4 80 0.6 0.1 0.2 15 0.2 0.1 0.2 5 0.2 0.2 0.4 20 Using the initial-rate method, what is the order of the reaction with respect to C? a. zero-order b. first-order c. third-order d. second-order e. impossible to tell from the data given


The dependence of the power of the reaction rate on the concentration is called the order of the reaction. The order of the reaction is the first order.

What is the initial-rate method?

The initial rate method is the estimation of the order of the reaction by the initial rates of the reactants and products and by performing the reaction several times by measuring the rate.

The reaction is given as,

\rm A + B + C \rightarrow Products

The rate of reaction can be given as:

\rm rate = k[A]^(x)[B]^(y)[C]^(z)

Here the variables x, y and z are orders respective to the reactant concentration and k is the rate constant.

Value of x with respect to A:

\begin{aligned} \rm \frac {Rate 3}{Rate 4} &= \rm [([A(3)])/([A(4)])]^(\rm x)\n\n(15)/(5) &= [([0.6])/([0.2])]^(\rm x)\n\n\rm x &= 1\end{aligned}

Value of y with respect to B:

\begin{aligned}\rm  \frac {Rate 2}{Rate 5} &= \rm [([B(2)])/([B(5)])]^(\rm y)\n\n(80)/(20) &= [([0.4])/([0.2])]^(\rm y)\n\n\rm y &= 2\end{aligned}

Value of z  with respect to C:

\rm \frac {Rate 1}{Rate 2} &= [([A(1)])/([A(2)])]^(x)  [([B(1)])/([B(2)])]^(y)  [([C(1)])/([C(2)])]^(z)

Substituting value of x = 1 and y = 2 in the above equation:

\begin{aligned}(160)/(80) &= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2) [([0.2])/([0.4])]^(\rm z)\n\n1 &= (0.5)^(\rm z)\n\n&= 1\end{aligned}

Therefore option b. with respect to C = 1, the order of the reaction is first-order.

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B. First order, Order with respect to C = 1


The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^(x)[B]^(y)[C]^(z)

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

(Rate3)/(Rate4)= [([A(3)])/([A(4)])]^(x)

(15)/(5)= [([0.6])/([0.2])]^(x)

3 = 3^(x) \n\nx =1

Order w.r.t B : Use trials 2 and 5

(Rate2)/(Rate5)= [([B(2)])/([B(5)])]^(y)

(80)/(20)= [([0.4])/([0.2])]^(y)

4 = 2^(y) \n\ny =2

Order w.r.t C : Use trials 1 and 2

(Rate1)/(Rate2)= [([A(1)])/([A(2)])]^(x)[([B(1)])/([B(2)])]^(y)[([C(1)])/([C(2)])]^(z)

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

(160)/(80)= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2)[([0.2])/([0.4])]^(z)

1 = (0.5)^(z)

z = 1

Therefore, order w.r.t C = 1

A sample containing only carbon, hydrogen, and silicon is subjected to elemental analysis. After complete combustion, a 0.7020 g sample of the compound yields 1.4 g of CO2, 0.86 g of H2O, and 0.478 g of SiO2. What is the empirical formula of the compound?


Answer: The empirical formula of compound is C_4H_(12)Si.


Mass of Sample= 0.702 g

Mass of CO_2 = 1.4 g

Mass of H_2O = 0.86 g

Mass of SiO_2 = 0.478 g  

First we have to calculate moles ofCO_2, H_2O and SiO_2 formed.

1. Moles of CO_2=(1.4g)/(44g/mol)=0.032mol

Now , Moles of carbon == Moles of CO_2 = 0.032

2.  Moles of H_2O=(0.86g)/(18g/mol)=0.048mol​​​

Now , Moles of hydrogen = 2* Moles of H_2O =2* 0.048=0.096mol

3.  Moles of SiO_2=(0.478g)/(60g/mol)=0.008 mol

Now , Moles of silicon = Moles of SiO_2 = 0.008 moles

Therefore, the ratio of number of moles of C : H : Si is  = 0.032 : 0.096 : 0.008

For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C= (0.032)/(0.008)=4

For H =(0.096)/(0.008)=12

For Si=(0.008)/(0.008)=1

Thus, C: H: Si = 4 : 12 : 1

The simplest ratio represent empirical formula.

Hence, the empirical formula of compound is C_4H_(12)Si.