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-BARSIC- [3]
3 years ago
9

1. The speed limit on most city streets is (25 miles). How many meters per minute

Chemistry
1 answer:
irina1246 [14]3 years ago
3 0

Answer:

Speed = 670.56 \frac{m}{min}

Explanation:

Given

Speed = \frac{25 mi}{hr}

Required

Convert to meters per minutes

Speed = \frac{25 mi}{hr}

Start by converting the speed from miles to meters

1\ mile = 1609.34\ meters.

So, we have:

Speed = \frac{25 * 1609.34\ m}{hr}

Speed = \frac{40233.5\ m}{hr}

Next, we convert time from hours to minutes

1\ hour = 60\ minutes

So, we have:

Speed = \frac{40233.5\ m}{60\ min}

Speed = 670.56 \frac{m}{min}

<em>Hence, the equivalent of 25 miles per hour is 670.56 meters per minutes</em>

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Write the autoionization reaction for methanol, ch3oh.
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Autoionization Reactions are those reactions in which ions or molecules ionizes spontaneously without adding any external reagent.

For Example,
                    Autoionization of water.

                               H₂O  +  H₂O   ⇆   H₃O⁺  +  OH⁻

Autoionization reaction of Methanol is shown below,

4 0
3 years ago
To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance.
kap26 [50]

To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance. You need to pipet 0 mL of the diluted solution to have 0.00 mg of iron.

In spectrophotometry, to plot the calibration curve, you need to prepare solutions with known concentrations and measure their absorbance.

We have a standard iron solution with a concentration of 0.2500g/L of pure iron (C₁). We pipet 25.00mL (V₁) of this standard iron solution into a 500mL (V₂) volumetric flask and dilute up to the mark with distilled water.

We can calculate the concentration of the diluted solution (C₂) using the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\\\C_2 = \frac{C_1 \times V_1}{V_2}  = \frac{0.2500 g/L \times 25.00 mL}{500 mL} = 0.0125 g/L

Then, if we wanted to prepare the blank, that is, the solution that contains the same matrix but not the analyte, and whose concentration in iron is 0.00 mg/L, we wouldn't pipet any of the diluted solution.

To plot the calibration curve, you need to prepare iron solutions with known concentrations and measure their absorbance. You need to pipet 0 mL of the diluted solution to have 0.00 mg of iron.

Learn  more: brainly.com/question/24195565

8 0
2 years ago
How does activation energy affect a chemical reaction ?
PilotLPTM [1.2K]

Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) increases. Figure 3: Lowering the Activation Energy of a Reaction by a Catalyst.

hope this helps!

(:

3 0
3 years ago
Read 2 more answers
Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc
tatiyna

Answer:

  • The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
  • The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%  
  • The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = \frac{4 }{3} \pi r^2

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = \frac{a\sqrt{2} }{4}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3 = \frac{\pi *a^3\sqrt{2}}{24}

Number of atoms/unit cell = 4

Total volume of the atoms = 4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6} = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = \frac{a\sqrt{3} }{4}

Volume of each atom =\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3 =\frac{\pi *a^3\sqrt{3}}{16}

Number of atoms/unit cell = 2

Total volume of the atoms = 2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8} = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = \frac{a\sqrt{3} }{8}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3 = \frac{\pi *a^3\sqrt{3}}{128}

Number of atoms/unit cell = 8

Total volume of the atoms = 8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16} = 0.3401

= 0.3401  X 100% = 34.01%

4 0
3 years ago
Given the equation: HCI + Na2SO4 ---&gt; NaCl + H2SO4 Hint: you should balance the equation If you start with 2.0 g of HCl (hydr
Rasek [7]

Answer:

approx 2.45g

Explanation:

2HCl + Na2SO4 → 2NaCl + H2SO4

 2    :       1           :      2       :      1

0.05                                     0.025 (moles)

⇒ mH2SO4 = 0.025 × 98 = 2.45 g

3 0
3 years ago
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