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fiasKO [112]
2 years ago
5

What is a large area of air with the same temperature?.

Chemistry
1 answer:
Yuliya22 [10]2 years ago
6 0

Sorry this is late but it would be the air mass

Hope this halpes :))

You might be interested in
What best describes the bonding in a water molecule
finlep [7]

Answer:

An oxygen atom shares a single electron with each H

Explanation:

H:O:H

Covalent bond, so

An oxygen atom shares a single electron with each H.

8 0
2 years ago
What is the mass of the object?
galben [10]

Answer:

37.3

263.5

Explanation:

The scale measures hundreds of units, tens of units, units, and parts of units (1 decimal place.

Scale 1

Hundreds 0 * 100 = 0

Tens: 3 * 10 = 30

Units: 7 * 1 = 7

1/10 unit = 3* 0.1 = 0.3

Total 30 + 7 + 0.3 = 37.3

Scale 2

Hundreds 2 * 100 = 200

Tens: 6 * 10 = 60

Units: 3 * 1 = 3

1/10 unit = 5* 0.1 = 0.5

Total = 200 + 60 + 3 + 0.5 = 263.5

7 0
3 years ago
A jet flies over the ocean. A sound wave travels from the jet to the surface of the ocean. Almost none of the energy from the so
lutik1710 [3]
We need to see the diagram
6 0
3 years ago
Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc
jeka94

Answer:

T_f = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

T_i = 21.5°C

C_{calorimeter} = 6.15 kJ/°C

T_f = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate ×  \frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

=  0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter =  C_{calorimeter} (T_f-T_i} )

21.84 kJ =6.15 kJ/°C * (T_f-21,5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - 132.225 kJ

21.84 KJ + 132.225 kJ = (6.15 kJ/^0C * T_f)

154.065 kJ = (6.15 kJ/^0C * T_f)

T_f = \frac{154.065kJ}{6.15kJ/^0C}

T_f =25.05°C

4 0
3 years ago
A perfect cube of aluminum metal was found to weigh 20.00 g. The density of aluminum is 2.7 g/ml. What are the dimensions of the
denis23 [38]

Answer:

             Height  = 1.9493 cm

             Width =  1.9493 cm

             Depth  =  1.9493 cm

Solution:

Data Given:

                  Mass  =  20 g

                  Density  =  2.7 g/mL

Step 1: Calculate the Volume,

As,

                                        Density  =  Mass ÷ Volume

Or,

                                        Volume  =  Mass ÷ Density

Putting values,

                                        Volume  =  20 g ÷ 2.7 g/mL

                                        Volume  =  7.407 mL or 7.407 cm³

Step 2: Calculate Dimensions of the Cube:

As we know,

                                        Volume  =  length × width × depth

So, we will take the cube root of 7.407 cm³ which is 1.9493 cm.

Hence,

                                        Volume  =  1.9493 cm × 1.9493 cm × 1.9493 cm

                                        Volume  =  7.407 cm³

4 0
3 years ago
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