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Artist 52 [7]
4 years ago
11

How much energy is required to melt 35.4 g of gold?

Chemistry
1 answer:
kap26 [50]4 years ago
4 0

Answer:

To determine the amount of heat the gold has absorbed to melt, we simply multiply the mass of the block of ice to the heat of fusion of water which is given above. We calculate as follows:

Heat = 20.0 g (35.4 g)

Heat = 1290 J

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In the Bee movie what does normal mean in the hive?
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Answer:

Acting how bees act. I had that question. :3.

6 0
3 years ago
1,500 grams is equal to:<br> 1.5 kg<br> 1.5L<br> 1.5 cm<br> all of the above
Salsk061 [2.6K]
1kg is 1,000 grams plus .5 is 1.5 kg
8 0
3 years ago
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
3 years ago
A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mix
Alisiya [41]

Answer:

M_f=38.8\%

Explanation:

From the question we are told that:

Pressure P=747mmHg

Temperature T=298K

Volume V=11.1

Heat Produced Q=780kJ

Generally the equation for ideal gas is mathematically given by

 PV=nRT

 n= (747/760) *11.1/ (0.0821*298)

 n=0.446mol

Therefore

 x+y=0.446

 x=0.446-y .....1

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

 x(889) + y(2220) = 760 ...... 2

Comparing Equation 1 and 2 and solving simultaneously

 x=0.446-y .....1

 x(889) + y(2220) = 760 ...... 2

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Therefore

Mole fraction 0f Methane is mathematically given as

 M_f=\frac{x}{n}*100\%

 M_f=\frac{1.173}{0.446}*100\%

 M_f=38.8\%

7 0
3 years ago
What is the mass of oxygen that can be produced from 2.79 moles of lead(ll) nitrate
denis23 [38]

1.38 moles of oxygen

Explanation:

Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

The mole ration of Lead (II) nitrate to oxygen is 2: 1

Therefore 2.76 moles of  Lead (II) nitrate will lead to production of? moles of oxygen;

2: 1

2.76: x

Cross-multiply;

2x = 2.76 * 1

x = 2.76 / 2

x = 1.38

8 0
3 years ago
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