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steposvetlana [31]
2 years ago
9

Which of these elements could be formed at the anode when a molten salt is electrolysed?

Chemistry
2 answers:
snow_lady [41]2 years ago
8 0
Hm mybeeeee just mybe it’s c
Oksana_A [137]2 years ago
6 0

Answer: A-Copper.

Explanation:

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How many moles of oxygen are needed for the complete combustion of 29.2 grams of acetylene?
podryga [215]

Moles of Oxygen= 2.8075 moles

<h3>Further explanation</h3>

Given

29.2 grams of acetylene

Required

moles of Oxygen

Solution

Reaction(Combustion of Acetylene) :

2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)

Mol of Acetylene :

= mass : MW Acetylene

= 29.2 g : 26 g/mol

= 1.123

From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :

= 5/2 x mol C₂H₂

= 5/2 x 1.123

= 2.8075 moles

7 0
2 years ago
Given the following isotopes of strontium, calculate the average atomic mass.
vodka [1.7K]

Answer:

87.5198

Explanation:

(43.65 * 87.05) + (48.25 * 87.93) + (8.11 * 87.50) = 8751.98

8751.98 / 100 = 87.5198

3 0
2 years ago
URGENT PLEASE HELP!! Earth science!!
aev [14]

Answer:

URGENT PLEASE HELP!! Earth science!!

What stages of development are represented by the Group 2 and Group 3 stars? Describe these

groups in terms of their relative age, size, brightness, and temperature.

3 0
2 years ago
The rate constant for this first‑order reaction is 0.900 s − 1 at 400 ∘ C. A ⟶ products How long, in seconds, would it take for
mafiozo [28]

Answer:

first-order reaction is 0.300 s–1 at 400 °C.. ... (in seconds) would it take for the concentration of A to decrease from 0.900 M

Explanation:

yep thats all i know

3 0
3 years ago
A. Which reactant is the limiting reagent?
Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2

c. m_{Br_2}^{leftover}=6.6g

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Zn+Br_2\rightarrow ZnBr_2

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2

Thus, the leftover of bromine is:

m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g

Best regards!

8 0
2 years ago
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