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victus00 [196]
3 years ago
11

Compute 4.659×104−2.14×104. Round the answer appropriately.

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
6 0
261.976 which rounds to 261.98(if rounding to the hundredths.
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What’s Silicon’s melting point?
torisob [31]

Answer:

2,577°F

Also,

Boron: 3,769°F (2,076°C)

Neon: -415.5°F (-248.6°C)

Beryllium: 2,349°F (1,287°C)

8 0
3 years ago
Read 2 more answers
P-fluoroanisole reacts with sulfur trioxide and sulfuric acid. Draw the major product of this substitution reaction; if applicab
Ipatiy [6.2K]

Answer: (Structure attached).

Explanation:

This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.

The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.

Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).

The reaction occurs in three steps:

  1. The attack on the electrophile forms the sigma complex.
  2. The loss of a proton regenerates an aromatic ring.
  3. The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).

Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.

8 0
3 years ago
Helppppppppppppppppppp me
masya89 [10]
The answer is A.
A pure substance is pure, so it cannot be separated in most cases.
8 0
4 years ago
Which statement defines specific heat capacity for a given sample
avanturin [10]

The statement that defines the specific heat capacity for a given sample is the quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

<h3>What is specific heat capacity?</h3>

Specific heat capacity is the of heat to increase the temperature per unit mass.

The formula to calculate the specific heat is Q = mct.

The options are attached here:

  1. The temperature of a given sample is 1 %.
  2. The temperature that a given sample can withstand.
  3. The quantity of heat that is required to raise the sample's temperature by 1 °C1 °C (Kelvin).
  4. The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

Thus, the correct option is 4. The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

Learn more about specific heat capacity

brainly.com/question/1747943

#SPJ1

8 0
2 years ago
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
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