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NikAS [45]
4 years ago
14

Lodine has a density of 4.94 g

Chemistry
1 answer:
atroni [7]4 years ago
6 0

Answer:

41.2 lb/gal

Explanation:

Step 1: Given data

Density of iodine: 4.94 g/mL

Step 2: Convert the mass to grams

We will use the conversion factor 1 lb = 454 g.

4.94 g/mL × (1 lb / 454 g) = 0.0109 lb/mL

Step 3: Convert the volume to gal

We will use the conversion factors 1 gal= 3.785 L and 1 L = 1,000 mL.

0.0109 lb/mL × (1,000 mL/1 L) × (3.785 L/1 gal) = 41.2 lb/gal

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What is Ka for H2CO3(aq)=H (aq0 HCO3-(aq)
horrorfan [7]
The Ka for H_2CO_3=H^++HCO_3^- is approximately 10^{-3.6}
6 0
4 years ago
How many moles are in 8.90 X 10^24 atoms of Na?
mixas84 [53]

Answer:

1.48 x 10^47 mol of Na

Explanation:

8.90 x 10^24 atoms of Na (1 mol of Na/6.022 x 10^23 atoms of Na)=

1.48 x 10^47 mol of Na

5 0
3 years ago
how much energy has your body used, in joules, if your health device indicates that 450 calories were burned during your workout
Likurg_2 [28]

Answer:

Total energy consumed = 1,882.8 joules

Explanation:

Given:

Calories burned = 450 calories

Find:

Total energy consumed

Computation:

1 calorie = 4.184 joules

So,

450 calories = 4.184 × 450

450 calories = 1,882.8 joules

Total energy consumed = 1,882.8 joules

7 0
3 years ago
Name the only satellite which is known to have an atmosphere in the planet it orbits
Maurinko [17]

Answer:

Titan which orbits Saturn

Explanation:

6 0
4 years ago
You wish to make a buffer with pH 7.0. You combine 0.060 grams of acetic acid and 14.59 grams of sodium acetate and add water to
aleksandr82 [10.1K]

Answer:

The pH of the buffer is 7.0 and this pH is not useful to pH 7.0

Explanation:

The pH of a buffer is obtained by using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where pH is the pH of the buffer</em>

<em>The pKa of acetic acid is 4.74.</em>

<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>

<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>

<em />

Replacing:

pH = 4.74 + log [0.1779mol] / [0.001mol]

<em>pH = 6.99 ≈ 7.0</em>

<em />

The pH of the buffer is 7.0

But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,

this pH is not useful to pH 7.0

<em />

7 0
3 years ago
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