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3 years ago

Which of the following is used for chemical symbols today? a. drawings c. letters b. icons d. numbers

Chemistry

1 answer:

devlian [24]3 years ago

3 0

Answer:

Letters

Explanation:

For example, today we use the periodic table which is full of elements named with 1 or 2 letters. Like how Helium is He and Sodium is Na. Hope this helps!!!

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Pure water has a ph of 10 at 298 K. True or false

anyanavicka [17]

Answer:

false

Explanation:

6 0

3 years ago

How many atoms are in 9.35 moles of lithium?

Fudgin [204]

Answer:

3.45 moles Li contains 2.08 × 10 (to the power of)24 atoms .

Explanation:

The relationship between atoms and moles is:

1 mole atoms =

6.022 × 10 (to the power of)23

atoms

In order to determine how many atoms occupy a given number of moles, multiply the given moles by

6.022 × 10 (to the power of)23

atoms/mole

In the case of 3.45 moles lithium (Li):

3.45 mol Li × 6.022 × 10 (to the power of)23 atoms Li/ 1 mol Li =

2.08 × 10 (to the power of)24

atoms Li rounded to three

6 0

3 years ago

Consider the following unbalanced redox reactions. In each case, separate the whole reactions into half-reactions, balance the h

Marina86 [1]

Answer & Explanation:

(a)

$Fe^{2+} +NO_{3}^{-} => Fe{(OH)}_3 + N_2$

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺ ⇒ Fe(OH)₃

reduction

NO₃⁻ ⇒ N₂

Oxidation Half Reaction

(redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1: 2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + 30H₂0 + 2NO₃⁻ + 12H⁺ + 10e⁻ ⇒10Fe(OH)₃ + 30H⁺ + 10e⁻ +

N₂ + 6H₂O

10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂

this is the balanced reaction

REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq) E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) - 2e⁻ ⇌ N₂(g) + 18H⁺ E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂ E°cell = -2.8V

E°cell = E°red + E°ox

4 0

4 years ago

A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th

olga_2 [115]

Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

s/342.30 = moles of sucrose. Also,

g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.

1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

6 0

3 years ago

Which quantity is equivalent to 39 grams of LiF?

Nutka1998 [239]

I think that it is 1.5 mole it might not be

3 0

3 years ago

Read 2 more answers