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2 years ago

True or False: The products of a chemical

Chemistry

2 answers:

attashe74 [19]2 years ago

6 0

Answer:

true

Explanation:

Ber [7]2 years ago

4 0

The anwser would be True

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The box in the picture begins motionless, and then both forces are applied at the same time. Describe the resulting motion of th

vaieri [72.5K]

It seems that you have missed to attach the given image for us to answer this question, but anyway, here is the answer. Based on the picture of a box which is motionless, the resulting motion of the box would be option B. The box will move to the left. Hope this answer helps.

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2 years ago

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What type of compound is Sodium hydroxide?<br> A.acid<br> B.base<br> C.element<br> D.salt

morpeh [17]

The answer is B
b base

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2 years ago

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Find the mass, in grams, of 3.758 mol CH4. ( show work )

vladimir2022 [97]

Answer:

The mass of CH4 is 60, 29 grams.

Explanation:

We use the weight of the atoms C and H for calculate the molar mass:

Weight of CH4= weight C+ 4 x weight H= 12,01 g/mol +4 x 1,008g/mol=

Weight of CH4 =16, 042 g/mol

1molCH4-----16, 042grams

3,758 mol CH4--X= (3,758 mol CH4 x 16, 042 grams)/1 mol CH4=60,285836 grams

5 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

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2 years ago

An emerald gemstone is an impure form of the mineral beryl, Be3Al2Si16O18. What is the percentage of silicon in the mineral?

solong [7]

Answer:

b. 54.9%

Explanation:

An emerald gemstone has the formula Be₃Al₂Si₁₆O₁₈. We can find the mass of each element in 1 mole of Be₃Al₂Si₁₆O₁₈ by multiplying the molar mass of the element by its atomicity.

Be: 3 × 9.01 g = 27.03 g

Al: 2 × 26.98 g = 53.96 g

Si: 16 × 28.09 g = 449.4 g

O: 18 × 16.00 g = 288.0 g

Total mass = 818.4 g

The mass percentage of silicon is:

(449.4 g / 818.4 g) × 100% = 54.91%

4 0

2 years ago