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3 years ago

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

Chemistry

1 answer:

kap26 [50]3 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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Lead (ll) nitride+ammonium sulfate > lead(ll) sulfate + ammonium nitride

Alenkinab [10]

Answer:

$Pb_3N_2 + 3 (NH_4)_2SO_4\rightarrow 3 PbSO_4 + 2 (NH_4)_3N$

Explanation:

Let's rewrite the given word equation in its chemical balanced equation representation:

1. Lead(II) nitride is represented by lead, Pb, in an oxidation state of 2+, while nitride is a typical nitrogen anion with a state 3-. As a result, the lowest common multiple between 2 and 3 is 6, meaning 2 lead cations are needed to balance 3 nitrogen anions: $Pb_3N_2$ .

2. Ammonium sulfate consists of an ammonium cation with a 1+ charge and sulfate anion with a 2- charge, two ammonium cations needed: $(NH_4)_2SO_4$ .

3. Lead(II) sulfate would have one lead cation and one sulfate anion, as they have the same magnitude of charges with opposite signs: $PbSO_4$ .

4. Ammonium nitride would require three amonium cations to balance the nitride anion: $(NH_4)_3N$ .

Let's write the balanced equation:

$Pb_3N_2 + 3 (NH_4)_2SO_4\rightarrow 3 PbSO_4 + 2 (NH_4)_3N$

7 0

3 years ago

A ridged steel tank filled with 62.7L of nitrogen gas of 85.0L atm

mylen [45]

The final gas pressure : 175.53 atm

<h3>Further explanation</h3>

Maybe the complete question is like this :

A ridged steel tank filled with 62.7 l of nitrogen gas at 85.0 atm and 19 °C is heated to 330 °C while the volume remains constant. what is the final gas pressure?

The volume remains constant⇒Gay Lussac's Law

<em>When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature </em>

$\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

P₁=85 atm

T₁=19+273=292 K

T₂=330+273=603 K

$\tt P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{85\times 603}{292}\\\\P_2=175.53~atm$

8 0

2 years ago

Why does cesium have a positive 1 charge

Sidana [21]

Answer:

1+1=5

Explanation:

4 0

3 years ago

Put the basic steps of pyruvate oxidation in order.

Lunna [17]

Answer:

1. carboxyl group is removed from pyruvate and releases 1 co2 molecule and leaves 2 carbon molecules

2. 2-carbon molecule is oxidized and electrons lost in oxidation are picked up by nad+ to form Ndah

3. oxidized 2 carbon molecule is attached to coenzyme a to form ACETYL COA

Explanation:

5 0

2 years ago

The sheet of gold foil Rutherford used was 304 mm wide and 0.016 mm thick. What maximum length of gold foil could be made from 1

katrin [286]

The maximum length of gold foil could be 1171.7 cm made from 1.10 x 10³ g of gold.

<h3>What is the density?</h3>

The density of an object can be described as the mass per unit volume. The average density is equal to the total mass divided by its total volume.

The mathematical formula of the density of the material can be expressed as follows:

Density = Mass/Volume

The density of a substance is an intrinsic property as it doesn't depend on its size and the S.I. unit of the density is Kg/m³.

Given the width of the gold foil, w = 304 mm = 30.4 cm

The height or thickness of the foil, h = 0.016 mm = 0.0016 cm

The density of the gold foil, d = 19.3 g/cm³

The mass of the gold given, m = 1100 g

The volume of the foil = m/d = 1100/ 19.3 = 56.99 g/cm³

As we know that the volume of gold foil , V = l × w × h

l = V/(w × h)

l = 56.99/(30.4 × 0.0016)

l =1171.7 cm

Learn more about density, here:

brainly.com/question/15164682

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8 0

1 year ago