All categories

3 years ago

In what way would readings from a digital thermometer be preferable to those from a liquid-based thermometer

Chemistry

2 answers:

lana66690 [7]3 years ago

8 0

Once calibrated a digital thermometer is faster and easir to use.

charle [14.2K]3 years ago

7 0

A digital thermometer has less errors in reading temperatures because it displays the actual reading rather than approximating it in a liquid-based thermometer. Also, digital thermometer are advantageous to use in extreme temperature conditions because they can withstand in these type of conditions.

You might be interested in

Hi can u help me pls? I'm totally stuck . The natural source of acidity in rain water is _____.

kykrilka [37]

Answer-The correct option is option d with says all of the above.

Explanation- All three acids that are given combined together to form acid rain in which nitric and sulphuric acid are stronger acids present while carbonic acid is a weaker one.

The carbon dioxide admitted in air combines with water to form carbonic acid and gives a weak acidic nature to rainwater. Pollution in nature makes sulphur and nitrogen present in air react to form the stronger acids responsible for acid rain.

5 0

3 years ago

How many molecules are present in 4.5 moles of H2O?

Gekata [30.6K]

Answer:

Mole of the H2O = 4.5

Number of molecules =4.5 multipled by avogadro's number.

6 0

2 years ago

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

Answer: The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

Explanation:

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago

While you were "sweating" your chemistry test, water vapor evaporates from your body, absorbing 50,000 J of energy. (assume no

TiliK225 [7]

Answer:

$\boxed {\boxed {\sf B. \ 22 \ grams}}$

Explanation:

We need to use the formula for heat of vaporization.

$Q=H_{vap}*m$

Identify the variables.

The heat absorbed by the evaporating water is the latent heat of vaporization. For water, that is 2260 Joules per gram.

Q is the energy, in this problem, 50,000 Joules.

m is the mass, which is unknown.

$H_{vap}=2260 \ J/g\\Q=50,000 \ J \\$

Substitute the values into the formula.

$50,000 \ J=2260 \ J/g*m$

We want to find the mass. We must isolate the variable, m.

m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.

$\frac{50,000 \ J}{2260 \ J/g} =\frac{2260 \ J/g*m}{2260 \ J/g}$

$\frac{50,000 \ J}{2260 \ J/g} =m$

Divide. Note that the Joules (J) will cancel each other out.

$\frac{50,000 \ }{2260 \ g} =m$

$22.1238938 \ g =m$

Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.

$22 \ g \approx m$

The mass is about 22 grams, so choice B is correct.

3 0

2 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago