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Vaselesa [24]
3 years ago
7

I’m extremely stuck on this, please help

Chemistry
1 answer:
nataly862011 [7]3 years ago
5 0

Answer:

ice: 2,100 J/(kg·K)

water: 4,200 J/(kg·K)

water vapor: 2,000 J/(kg·K)

basalt: 840 J/(kg·K)

granite: 790 J/(kg·K)

aluminum: 890 J/(kg·K)

iron: 450 J/(kg·K)

copper: 380 J/(kg·K)

lead: 130 J/(kg·K)

Explanation:

Specific heat is the amount of thermal energy you need to supply to a sample weighing 1 kg to increase its temperature by 1 K.

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A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
What is the mass (grams) of salt in 10.0 m' of ocean water? ball park-4x10's (1.000 molsalt -58.44 g salt, 1.0 L ocean water -0.
koban [17]

Answer:

3.5 × 10⁵ g of salt

Explanation:

<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>

We have this data:

  • 1.000 mol salt is equal to 58.44 g salt
  • 1.0 L of ocean water contains 0.60 mol of salt

We will need the following relations:

  • 1 L = 1dm³
  • 1 m³ = 10³ dm³

We can use proportions:

10.0m^{3} .\frac{10^{3}dm^{3}  }{1m^{3} } .\frac{1L}{1dm^{3} } .\frac{0.60molSalt}{1.0L} .\frac{58.44gSalt}{1molSalt} =3.5 \times 10^{5} gSalt

8 0
3 years ago
What is the molar concentration of a solution prepared by dissolving 8.63 g of Ba(NO3)2 (molar mass = 261.35 g/mol) to a total v
PilotLPTM [1.2K]

Answer:

我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈

Explanation:

我實際上不知道我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈

7 0
3 years ago
Write the name and formula for the chemical substance which is produced in most acid-base neutralization reactions
Minchanka [31]
Acid + base gives you salt and H2O
7 0
3 years ago
Copper was the first metal to be produced from its ore because it is the easiest to smelt, that is, to refine by heating in the
Ganezh [65]

Answer:

57.48%

Explanation:

Calculate the mass of 1 mole of malachite:

MM Cu = 63.55

MM O = 16.00

MM H = 1.01

MM C = 12.01

(Cu_{2}(OH)_{2}CO_{3})

A mole of malachite has:

2 moles of Cu

5 moles of O

2 moles of H

1 mole of C

MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)

MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01

MW Malachite = 221.13

Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1

Now divide the mass of Cu by the mass of Malachite

%Cu = \frac{127.1}{221.13} =0.5748=57.48%

7 0
3 years ago
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