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earnstyle [38]
3 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]3 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]3 years ago
4 0
I guess B is the right one......
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7.6 The diagrams show the atoms in four different substances. Each circle represents an atom.
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Answer:

A and C represent elements while B and D represent Compounds

Explanation:

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1 year ago
Is this right? giving brainlt if you actually checked​
Maksim231197 [3]

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5 0
2 years ago
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When ammonium chloride is dissolved in water, the beaker gets cold. Which type of reaction does this describe
gulaghasi [49]

Answer:

B

Explanation:

i think with exothermic reactions heat is released

this is what i looked up

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7 0
3 years ago
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.
Hitman42 [59]

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = 2 \times 0.475

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = 2 \times 0.025

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = pK_{a} + log \frac{Base}{Acid}  

              = 4.2 + log \frac{0.05}{0.045}

              = 4.245

For,  

         HCOOH + NaOH \rightarrow HCOONa + H_{2}O

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = pK_{a} + log \frac{Base}{Acid}  

          4.245 = 3.75 + log \frac{Base}{Acid}

      log \frac{Base}{Acid} = 0.5

    \frac{Base}{Acid} = 3.162

Now,

        \frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

                 

8 0
3 years ago
you have accidentally broken a test tube and spilled a chemical on the bench. which of the following best describes what you sho
ANEK [815]

Tell the teacher, do NOT clean it up yourself.

4 0
2 years ago
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