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pav-90 [236]
3 years ago
13

what happens to the density of an object when the volume of that object increases and the mass remains the same?

Chemistry
1 answer:
Dima020 [189]3 years ago
6 0

Answer:

d=m/v       If v increases then the density decreases.   Denominator get  larger

Explanation:

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3. The exosphere is the outer layer of the
barxatty [35]

Answer:

D. exosphere is the outer layer of the thermosphere

3 0
3 years ago
Please help me, I really don't want to fail but I don't know how to do this
Brrunno [24]

Answer:

A)

<u>4, 7, 4, 6</u>

B)

<u>12 moles</u>

Explanation:

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

__↑______↑

8.00 mol | 14.00 mol

________________

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

(Reactant = Product)

Reactant: 1a [N] Product: 1c.

Reactant: 3a [H] Product: 2d.

Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

3a = 2d

2b = 2c + 1d.

Then just use the substitution methods to solve.

3 0
2 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
4 Which pair of ions is isoelectronic?
Softa [21]

Answer:

B. Na+ and O2-

Explanation:

Na+ plus has 10 electrons and O2- also has 10 electrons

7 0
2 years ago
A balloon moves vertically upward at
Usimov [2.4K]

Answer:

B. Velocity

Explanation:

The answer is Velocity because the def of velocity is the rate of constant speed in a given direction. I hope this helps, have a good night.

6 0
3 years ago
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