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S_A_V [24]
3 years ago
12

The heat change that accompanies the formation of 1.00 mol of carbon dioxide from its elements is −393.7 kJ/mol. What heat chang

e accompanies the formation of 0.631 mol CO2?
Chemistry
1 answer:
gladu [14]3 years ago
7 0

Answer:

-248.425 kJ/mol

Explanation:

This question can be done by simply taking the ratios:

Since the given the heat change for 1.00 mol of CO_2 is -373.7 KJ/mol. Let us consider the required Heat Exchange at 0.631 mol of

We can write:

\frac{-373.7} {1.00} = \frac{x}{0.631}

x = ( -373.7)(0.631)

x = -248.425 KJ/mol

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For the reaction A(g)
zubka84 [21]

Answer:

Yes

Explanation:

By definition, the equilibrium constanct, Kc, for the reaction A ⇒ 2B is

= [A]^1 / [B]^2

Substitute [A] = 4 and [B] = 2 in the equation,

[A]^1 / [B]^2

= 4^1 / 2^2

= 1

= Kc

So yes the reaction is at equilibrium.

6 0
2 years ago
Does pb(no3)2 + Na3(PO4) = Pb3(PO4)2 + Na(NO3) have a precipitate?
ipn [44]

Answer:

Yes, Pb3(PO4)2.

Explanation:

Hello there!

In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.

Regards!

4 0
3 years ago
1. Compare masses: a) 0,4mol CO₂ and 0,6mol H₂O ; b) 0,135mol H₂SO₄ and 0,5mol HCI.
bixtya [17]

Answer:

you can now deduct which one is greater or smaller and by how much.

Explanation:

no of moles= mass/molar mass

1ai) 0.4 = m/ ( 12 + (16*2)

m= 0.4* 44

m= 17.6g

ii) 0.6= m/( 2*1 + 16)

m= 0.6 *18

m= 10.8g

b) 0.135 = m/ ( 2*1 +32 + (16*4)

m= 0.135* 98

m= 13.23g

ii) 0.5= m/ (1+35.5)

m= 0.5*36.5

m= 18.25g

2. Avogadro's Number = 6.02×10²³

1 mol of any element= 6.02×10²³ particles

a) 0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules

ii) 0.3 mol of CO2= (6.02×10²³) * 0.3

= 1.806 × 10²³ molecules

Ans: 0.3 mol of CO2

bi) 0.25 mol of HCl= (6.02×10²³) * 0.25

= 1.505 × 10²³ molecules

bii) - find the no of moles first:

no of moles= mass/molar mass

n= 3.4g/ 34g →mr of H2S in g=2+32= 34g

n= 0.l mol

- use the Avogadro Number.

0.1 mol of H20= (6.02×10²³) * 0.1

= 6.02×10²² molecules.

biii) here you're given the density, use it to find the mass of acetic acid.

ρ = 1049 g/ml

ρ = m/v, where v=5 ml

1049 = m/ 5

m= ρ*v

m= 1049*5

m= 5245 g

• convert this into moles.

mr of CH3COOH= 12 + 3+ 12+ 16+ 16+ 1

= 60

mr in g = 60g

n= m/mr

n= 5245/ 60

n= 87. 41666...

n= 87.4 moles

•using Avogadro's Number:

87.4 moles of acetic acid=(6.02×10²³)*87.4

= 2.25148* 10²⁵

= 2.25 * 10²⁵ molecules

thus, the ans for this is 5 ml of acetic acid.

7 0
2 years ago
Platinum crystallizes with the face-centered cubic unit cell. The radius of a platinum atom is 139 pm. Calculate the edge length
horrorfan [7]

Answer:

Length = 393pm, Density = 21.3 g/cm^3.

Explanation:

From the question above, we have the following parameters or data which is going to aid in solving the above Question.

=> The radius of a platinum atom = 139 pm.

Therefore, the length can be calculated by making use of the formula given below:

Length = 2 √( 2r) = 2 × √ (2 × 139 × 10^-12m ) = 393 × 10^-10 m = 393pm.

The density can be calculated by making use of the chemical formula given below:

Density = mass ÷ volume = (195.064/ 6.02 × 10^23) ÷ (3.93 × 10^-10/ 10^-2) = 21.3 g/cm^3.

3 0
3 years ago
6. A piece of solid gold was heated from 274K to 314K. 35.73 of energy was needed to raise the temperature.
ivanzaharov [21]

From Q = mcΔT, we can rearrange the equation to solve for mass, m = Q/cΔT. The specific heat capacity, c, of solid gold is 0.129 J/g °C. I'm assuming that the energy is given in joules, as it's not specified in the question as written.

m = Q/cΔT = (35.73 J)/(0.129 J/g °C)(40.85 °C - 0.85°C)

m = 6.92 g of gold was present  

5 0
3 years ago
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