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Furkat [3]
4 years ago
9

Write these numbers rounded to the three nearest significant figures

Chemistry
1 answer:
Savatey [412]4 years ago
7 0

Answer:

(a) 52.2 mL helium

(b) 18.0 g nitrogen

(c) 78.5 mg MSG

(d) 23,600,000 mm wavelength

(e) 0.00420 kg lead

Rounded the numbers to three nearest significant figures.

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The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia
Sonja [21]

Answer:

The answer is "\bold{50.42 \pm 0.08}".

Explanation:

Overall delivered volume = [(49.06 \pm 0.05) + (1.77 \pm 0.05)]\ mL

Its blank solution without any of the required analysis  = (0.41 \pm 0.04)\ mL

Compute the volume of the endpoint as follows:  

Formula:

\text{End point volume = Total Volume delivered - volume required}

= (49.06 \pm 0.05) + (1.77 \pm 0.05) - (0.41 \pm 0.04) \\\\= (49.06 + 1.77 - 0.41) \pm  \ \ (absolute \ \  uncertainty)

therefore,

absolute uncertainty =\sqrt{(0.05)^2 + (0.05)^2 + (0.04)^2}

                                   =\sqrt{0.0025 +0.0025 +0.0016} \\ \\=\sqrt{0.0066}\\\\=0.08124\\

The Endpoint volume = (49.06+1.77-0.41)\pm (0.08124)

                                    = 50.42 \pm 0.08

Therefore, the volume of the endpoint adjusted for the blank is:

\bold { = 50.42 \pm 0.08}

3 0
4 years ago
Which type of polymer is RNA?
Alex787 [66]
a nucleic acid
Explanation: The nucleic acids, both DNA and RNA, consist of polymers ofnucleotides. The nucleotides are linked covalently between the 3' carbon atom of the pentose and thephosphate group attached to the 5' carbon of the adjacent pentose
5 0
4 years ago
Read 2 more answers
Calculate the pH when 64.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)
KonstantinChe [14]

Answer:

The answer is "12.06"

Explanation:

Given:

M(HBrO) = 0.3\ M\\\\V(HBrO) = 20 \ mL\\\\M(KOH) = 0.15 \ M\\\\V(KOH) = 64 \ mL

\to mol(HBrO) = M(HBrO)  \times  V(HBrO) = 0.3 M   \times 20 mL = 6 \ mmol\\\\\to mol(KOH) = M(KOH)  \times   V(KOH)= 0.15 M  \times  64 mL = 9.6 mmol

  6 mmol of both will react

excess KOH remaining= 3.15 \ mmol

Volume= 20 + 64 = 84 \ mL

[OH^{-}] = \frac{ 9.6 \ mmol}{84\  mL} = 0.01142\ M

use:

pOH = -\log [OH^-]

        = -\log (1.142\times 10^{-2})\\\\= 1.94

use:

PH = 14 - pOH

       = 14 - 1.94\\\\= 12.06

4 0
3 years ago
What is the density of a bar of lead that weighs 173 g and has the following dimensions: l= 2.00 cm, w = 3.00 cm, h = 1.00 in?
oksano4ka [1.4K]

Answer:

\rho=7373.2g/cm^3

Explanation:

Hello!

In this case, since the density is defined as the degree of compactness of a substance, and is defined as the mass over the volume:

\rho= \frac{m}{V}

Given the dimensions of the bar of lead, we can compute the volume as shown below:

V=2.00cm*3.00cm*1.00in*\frac{1in}{2.54cm} \\\\V=2.36cm^3

Thus, the density turns out:

\rho=\frac{173g}{2.36cm^3} \\\\\rho=7373.2g/cm^3

Best regards!

8 0
3 years ago
Which element requieres the least amount of energy to remove the outermost electron
Arte-miy333 [17]

Answer: Sodium

Explanation: The sodium has only one electron at the outer layer.

Hope this helps! :)

8 0
4 years ago
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