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trasher [3.6K]
3 years ago
8

This is the picture answer this plsss..

Chemistry
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

he is a weelder because of him taking a peice of metal and recreateing it.

Explanation:

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Calculate AS° for the reaction below: N2(g)+202(g) 2NO2(g) where ASo for N2(g), O2(g), & NO2(g), respectively, is 191.5, 205
nignag [31]

Answer:

ΔS Rx = -120, 5 J/K

Explanation:

The ΔS in a reaction is defined thus:

ΔS Rx = ∑ n S°products - ∑ m S°reactants

For the reaction:

N₂(g) + 2 O₂(g) → 2NO₂(g)

ΔS Rx = 2 mol × 240,5 \frac{J}{mol.K} - [ 1 mol × 191,5 \frac{J}{mol.K} + 2 mol × 205,0 \frac{J}{mol.K}]=

<em>ΔS Rx = -120, 5 J/K</em>

A negative value in ΔS means a negative entropy of the process. Doing this process entropycally unfavorable.

I hope it helps!

5 0
3 years ago
How much heat is released as 5.00G of PB cool from 75.0° C to 25.0° C
zalisa [80]

Answer:

32.25 J

Explanation:

Heat, H = ?

Mass, m = 5g = 5 / 1000 = 0.005 Kg

Initial Temperature = 75.0° C

Final Temperature = 25.0° C

Temperature change, ΔT = Final - Initial = 50° C

Specific heat capacity of lead, C = 129 (J/kg°C)

The relationship between these quantities is given by the equation

H = mCΔT

H = 0.005 * 129 * 50

H = 32.25 J

7 0
3 years ago
element x is a neutral element and has 39 protons. what element is it and how many electrons does it have? (use the periodic tab
alisha [4.7K]

The element itself is Yttrium, which is a transition metal and not that reactive. Since it has 39 protons it will also have 39 electrons.

4 0
3 years ago
Did you write these compounds?<br> KBr<br> 2 NH₂OH<br> 2HNO3
Orlov [11]

The missing components of the neutralisation reaction include the following:

  • KBr
  • KBr 2NH4OH
  • KBr 2NH4OH2HNO2

<h3>What is neutralisation reaction?</h3>

Neutralisation reaction is defined as the type of reaction that leads to the formation of salt and water when an acid and a base reacts.

From the reactions given the missing components are replaced as follows:

  • HBr + KOH --> KBr + H2O

  • H2SO4 + 2NH4OH --> (NH4)2SO4 + 2H2O

  • 2HNO3 + Mg(OH)2 --> Mg(NO3)2 + 2H2O

Learn more about acids here:

brainly.com/question/26353151

#SPJ1

7 0
2 years ago
Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?
WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V  

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

6 0
3 years ago
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