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2 years ago

Which is the best reason for using a temperature probe instead of a simple laboratory thermometer?

Chemistry

2 answers:

LenaWriter [7]2 years ago

6 0

Answer:

A temperature probe/sensor is more accurate than a regular lab thermometer.

Explanation:

Since a temperature probe reads heat emission using infrared, it is a lot more accurate and precise than a regular thermometer, which uses physical touch.

bixtya [17]2 years ago

5 0

Answer: to record continuous measurements over several minutes

Explanation:

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A 25.0-ml sample of hno3 solution requires 35.7 ml of 0.108 m na2co3 to completely react with all of the hno3 in the solution. w

Rasek [7]

If we write the equation of the reaction that will take place, it is:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂CO3

The molar ratio of 2HNO₃ : Na₂CO₃ = 1 : 2
Therefore, we can set up the equation:
M₁V₁ = 2M₂V₂
Where the left side of the equation has the molarity and volume of HNO₃ and the right side has the molarity and concentration of Na₂CO₃. Substituting:
M₁ = (2 x 0.108 x 35.7) / 25
M₁ = 0.308 M

5 0

2 years ago

A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav

dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows:

$Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles$

$Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles$

$_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e$

By the stoichiometry of above reaction,

1 mole of $_{40}^{91}\textrm{Zr}$ is produced by 1 mole $_{41}^{91}\textrm{Nb}$

So, 0.17 moles of $_{40}^{91}\textrm{Zr}$ will be produced by = $\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}$

Amount of $_{82}^{212}\textrm{K}$

decomposed will be = 0.17 moles

Initial amount of $_{40}^{91}\textrm{Nb}$ will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.024

$a_o$ = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

$0.024=\frac{0.194}{2^n}$

$n=3$

Therefore, 3 half lives have passed.

3 0

2 years ago

Read 2 more answers

The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Use Ideal Gas Law (PV =

Scorpion4ik [409]

Answer:

26.3 mL

Explanation:

Step 1:

Obtaining an appropriate gas law from the ideal gas equation.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by T

PV/T = nR

At this stage, we'll assume the number of mole (n) to be constant.

Note: R is the gas constant.

PV/T = constant.

We can thus, write the above equation as:

P1V1/T1 = P2V2/T2

The above equation is called the general gas equation.

Step 2:

Data obtained from the question. This includes the following:

Initial volume (V1) = 27.5 mL

Initial temperature (T1) = 22.0°C = 22.0°C + 273 = 295K

Initial pressure (P1) = 0.974 atm.

Final temperature (T2) = 15.0°C = 15.0°C + 273 = 288K

Final pressure (P2) = 0.993 atm

Final volume (V2) =..?

Step 3:

Determination of the final volume of the gas using the general gas equation obtained. This is illustrated below:

P1V1 /T1 = P2V2/T2

0.974 x 27.5/295 = 0.993 x V2/288

Cross multiply to express in linear.

295x0.993xV2 = 0.974x27.5x288

Divide both side by 295 x 0.993

V2 = (0.974x27.5x288)/(295x0.993)

V2 = 26.3 mL

Therefore, the new volume of the gas is 26.3 mL

4 0

2 years ago

Suppose there is a gaseous mixture of nitrogen and oxygen. If the total pressure of the mixture is 470 mmHg , and the partial pr

ch4aika [34]

Answer:

190 mmHg

Explanation:

According to Dalton's law, in a mixture of ideal gases, each gas behaves independently of the other. Also, the total pressure is equal to the sum of the individual partial pressures.

The total pressure of the mixture is 470 mmHg , and the partial pressure of nitrogen is 280 mmHg. Then,

P = pO₂ + pN₂

pO₂ = P - pN₂

pO₂ = 470 mmHg - 280 mmHg

pO₂ = 190 mmHg

5 0

2 years ago

Consider the picture of a gas pump. Which type of gasoline has the highest percentage of octane (the main component of gasoline)

Readme [11.4K]

Answer:

premium: 91 octane rating

Explanation:

Octane number refers to the percentage or volume fraction of isooctane in a fuel.

The octane number gives a picture of how safe a fuel is for an engine. The higher the octane rating the lesser the tendency of the fuel to cause knocking of the engine.

The type of gasoline with the highest percentage of octane among the options is premium.

5 0

2 years ago

Read 2 more answers