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earnstyle [38]
4 years ago
6

Many types of scientific equipment are used to perform different functions in the science lab. Which of the following combinatio

ns of equipment would be needed to bring 1 L of water to 85°C
A. Graduated cylinder, balance, beaker and hot plate
B. Graduated cylinder, beaker, thermometer and hot plate
C. Test tube, thermometer, and Bunsen burner
D. Test tube, electronic balance and a hot plate
Chemistry
2 answers:
Minchanka [31]4 years ago
7 0
To heat up water we need hot plate. 
To measure the temperature we need Thermometer. 
To measure 1L of water we need Graduated Cylinder. 
And we need a container to heat the water: Beaker
So the answer is B
Phantasy [73]4 years ago
4 0
I guess B is the right one......
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A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3
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Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of Pb(IO_{3})_{2} is given as follows-

                   Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}

Hence, if solubility of Pb(IO_{3})_{2} is S (M) then-

                             [Pb^{2+}]=S(M) and [IO_{3}^{-}]=2S(M)

Where species under third bracket represent equilibrium concentrations

So, solubility product of Pb(IO_{3})_{2} , K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}

Here, [Pb^{2+}]=S(M)=5.0\times 10^{-5}M

So, [IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M

So, K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}

Hence option (E) is correct

7 0
3 years ago
a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp
Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

Learn more about heat of solution:brainly.com/question/24243878

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