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Lelechka [254]
3 years ago
14

Use exponential notation to express the number 12,500 to

Chemistry
1 answer:
Eva8 [605]3 years ago
6 0

1

2

5

Explanation:

start from the left tor right

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Which statement about anabolism is true?
skelet666 [1.2K]

Answer:

i would say B? Im sorry if im wrong

Explanation:

i think that because Catabolic reactions release energy, while anabolic reactions use up energy. Anabolism is the opposite of catabolism.

8 0
3 years ago
Read 2 more answers
A 0.6467-g portion of manganese dioxide was added to an acidic solution in which 1.1701 g of a chloride-containing sample was di
Katen [24]

Answer:

29.39% of AlCl₃ in the sample

Explanation:

Based on the reaction:

MnO₂(s) + 2Cl⁻ + 4H⁺ → Mn²⁺ + Cl₂(g) + 2H₂O

We can find the amount of chloride in solution with the amount of MnO₂ that reacted as follows:

<em>Initial mass MnO₂ = 0.6467g</em>

<em>Recovered mass = 0.3104g</em>

Mass that reacted = 0.6467g - 0.3104g = 0.3363g

<em>Moles MnO₂ -Molar mass: 86.9368g/mol-:</em>

0.3363g * (1mol / 86.9368g) = 3.868x10⁻³ moles MnO₂

<em>Moles Cl⁻:</em>

3.868x10⁻³ moles MnO₂ * (2mol Cl⁻ / 1mol MnO₂) = 7.737x10⁻³ moles Cl⁻

<em>Moles of AlCl₃ and mass -Molar mass AlCl₃: 133.34g/mol-:</em>

7.737x10⁻³ moles Cl⁻ * (1mol AlCl₃ / 3mol Cl⁻) = 2.579x10⁻³ moles AlCl₃

2.579x10⁻³ moles AlCl₃ * (133.34g / mol) =

<em>0.3439g of AlCl₃</em> are present in the sample.

The percent is:

0.3439g of AlCl₃ / 1.1701g * 100 =

<h3>29.39% of AlCl₃ in the sample</h3>

7 0
2 years ago
Which of the following you do NOT need to pay attention while using a centrifuge?
Phoenix [80]

Answer:

The radius of the centrifuge.

Explanation:

Hello,

Since the radius of the centrifuge is just a design parameter, it wouldn't be a cause of failure because it is used to know how many tubes could be fitted in into the centrifuge. On the other hand, keeping you attention away from other factors could turn into a failure as long as the sample could be poured down or just turn out inadequate for the expected results.

Best regards.

8 0
3 years ago
At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements
Gnom [1K]

Given that:

  • At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

     Initial volume of    Final volume of    Mass of water   Density (g/mL)

     burette (mL)        burette   (mL)       dispensed (g)

 

Sample 1      2.33                     7.34                   5.000               -----

Sample 2      7.34                    12.37                 5.025                -----

Sample 3      12.37                   18.50                6.112                  -----

Sample 4      18.50                  24.57               6.064                 -----

Sample 5     24.57                  31.31                6.720                  -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

  • The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

\mathbf{density = \dfrac{mass}{volume}}

\mathbf{density = \dfrac{5.01 g}{5.000 ml}}

density = 0.998004 g/ml

Sample 2:

\mathbf{density = \dfrac{5.025 g}{5.03ml}}

density = 0.999006 g/ml

Sample 3:

\mathbf{density = \dfrac{6.112 g}{6.13ml}}

density = 0.997064 g/ml

Sample 4:

\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}

density = 0.999012 g/ml

Sample 5:

\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}

density = 0.997033 g/ml

Thus, the average density for all the samples is:

\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 +   0.999012  + 0.997033  )}{5}}

= 0.998024

∴

The percentage error for the two densities measurement is:

=\dfrac{ (experimental \  value -theoretical  \ value)\times 100 }{theoretical  \ value}

Given that the theoretical value = 0.997044 g/ml

Then;

\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

4 0
2 years ago
What is the ph of a solution that results from mixing 25 ml of 0.15 M hcl to 25 ml of 0.52 m nh3?
BartSMP [9]

Answer:

7.00

Explanation:

When the solutions are mixed, the HCl dissociates to form the ions H+ and Cl-. The ion H+ will react with the NH3 to form NH4+. The stoichiometry for this is 1 mol of HCl to 1 mol of H+ to 1 mol of Cl-, and 1 mol of H+ to 1 mol of NH3 to 1 mol of NH4+.

First, let's find the number of moles of each one of them, multiplying the concentration by the volume:

nH+ = 0.15 M * 25 mL = 3.75 mmol

nNH3 = 0.52 M * 25 mL = 13 mmol

So, all the H+ is consumed, and the neutralization is completed, thus pH will be the pH of the solvent (water), pH = 7.00.

5 0
3 years ago
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