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andrezito [222]
3 years ago
13

Please help I will reward Brainly

Chemistry
2 answers:
Leni [432]3 years ago
8 0
The answer is first option.
Lera25 [3.4K]3 years ago
4 0
The first option: the basic unit of the nervous system
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How many moles of iron are there in 55.85g of Fe3O4
soldier1979 [14.2K]

Answer:

• Molecular mass of Iron (III) tetraoxide

\dashrightarrow \: { \tt{(56 \times 3) + (16 \times 4)}} \\  = { \tt{168 + 64}} \\  = { \tt{232\:g}}

[ molar masses: Fe → 56, O → 16 ]

\dashrightarrow \:{ \rm{232 \: g \:  = 1 \: mole}} \\ \\   \dashrightarrow \: { \rm{55.85 \: g = ( \frac{55.85}{232}) \: moles }} \\  \\ \dashrightarrow \:  { \boxed{ \tt{ = 0.24 \: moles}}}

8 0
2 years ago
PLEASE HELP <br> I do not understand
Vikentia [17]

Answer:

Absorbing beta particle because the beta is the numbers and are less and the  big numbers are positive and they are the alpha so when you add beta particle it is called Absorbing so the answer is Absorbing beta particle

4 0
2 years ago
Read 2 more answers
How many moles of copper are there in 6.93 g of copper sample<br>pls be quick​
Alex_Xolod [135]
  • Molar mass of copper=63.5g/mol
  • Given mass=6.93g

\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}

\\ \sf\longmapsto No\:of\:moles=\dfrac{6.93}{63.5}

\\ \sf\longmapsto No\:of\:moles=0.109\approx 1.11moles

7 0
2 years ago
Read 2 more answers
What is the ph of the benzoic acid solution prior to adding sodium benzoate?
Reika [66]
The acid dissociation constant of benzoic acid is 6.5 x 10^-5. Therefore, the pH of the benzoic acid solution prior to adding sodium benzoate is:

pH = -log[Ka]
pH = -log (6.5 x 10^-5)
pH = 4.19

The pH of the benzoic acid solution is 4.19 which is acidic, but a weak acid. 

7 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
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