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Veronika [31]
2 years ago
10

why do you use a graduated cylinder to measure out the desired volume of koh and h2so4, rather than a pipet or a buret?

Chemistry
1 answer:
sergey [27]2 years ago
7 0

The graduated cylinder is used to measure the volume of KOH and H2SO4 when accurate volume measurement is not required.

In the laboratory certain graduated apparatus are used to measure liquids. These graduated apparatus used to measure liquids include;

  • burette
  • pipette
  • measuring cylinder
  • Erlenmeyer flask

Sometimes, we are not really looking for a strictly accurate volume of liquid and we can use a graduated cylinder to measure the volume of liquid in such cases.

However, when we need to have strictly accurate volume measurement, we need a pipet or a buret.

Learn more: brainly.com/question/15670537

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Find the molecular mass of<br>MgSO4 7H₂O.​
mario62 [17]

Answer:

MgO +H2SO4

Explanation:

use photomath

7 0
3 years ago
A scientist performs a cutting-edge experiment with exciting results. What
algol [13]

Answer:

B

Explanation:

It's B because the first trial of an experiment may not always be right so you want to run multiple trials

4 0
3 years ago
Which of the following is a molecular formula for a compound with an empirical formula of CH2O and a molar mass of 150. g/mol.
trapecia [35]

Answer:

C₅H₁₀O₅

Explanation:

Let's consider a compound with the empirical formula CH₂O. In order to determine the molecular formula, we have to calculate "n", so that

n = molar mass of the molecular formula / molar mass of the empirical formula

The molar mass of the molecular formula is 150 g/mol.

The molar mass of the empirical formula is 12 + 2 × 1 + 16 = 30 g/mol

n = (150 g/mol) / (30 g/mol) = 5

Then, we multiply the empirical formula by 5.

CH₂O × 5 = C₅H₁₀O₅

7 0
3 years ago
A pool is 60.0 m long and 35.0 m wide. How many mL of water are needed to fill the pool to an average depth of 6.35 ft? Enter yo
ra1l [238]

Answer:

4.07 × 10⁹ mL

Explanation:

Step 1: Given data

Length of the pool (L): 60.0 m

Width of the pool (W): 35.0 m

Depth of water in the pool (D): 6.35 ft

Step 2: Convert "D" to m

We will use the relationship 1 m = 3.28 ft

6.35ft \times \frac{1m}{3.28ft} = 1.94m

Step 3: Calculate the volume of water (V)

We will use the following expression.

V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}

Step 4: Convert "V" to mL

We will use the relationship 1 m³ = 10⁶ mL.

4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL

7 0
3 years ago
What is 72.9 g HCI into moles
meriva
1.9993999057621666 moles
1 mole = 36.46094
6 0
3 years ago
Read 2 more answers
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