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Karolina [17]
3 years ago
10

Liquid to gas require heat?

Chemistry
2 answers:
serg [7]3 years ago
7 0

Answer:

yes

Explanation:

let's say you have ice, you put it on a pan then it heats up you get a liquid after a while it turns into a gas

Anton [14]3 years ago
5 0

Answer:

yes

Explanation:

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Some elements are solids at room temperature true or false
nadya68 [22]
This is true, think of silver, gold, etc— those are heated to make them liquid
7 0
3 years ago
If a 60-g object has a volume of 30 cm3, what is its density? 2 g/cm3 0.5 cm3/g 1800 g * cm3 none of the above
jasenka [17]

Answer : The density of an object is, 2g/cm^3

Solution : Given,

Mass of an object = 60 g

Volume of an object = 30cm^3

Formula used :

\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}

Now put all the given values in this formula, we get the density of an object.

\text{Density of an object}=\frac{60g}{30cm^3}=2g/cm^3

Therefore, the density of an object is, 2g/cm^3

6 0
3 years ago
Please help asap loll
BlackZzzverrR [31]

Explanation:

Metals -

  1. Left side of periodic table
  2. Malleable
  3. Ductile
  4. Good conductor
  5. Mostly solid
  6. Shinly lusture

Non-metals -

  1. Poor conductor
  2. Dull
  3. Brittle
  4. Right side of periodic table

Metalloids -

  1. Properties of metals and non-metals.
  2. Found on the 'staircase' of the periodic table.

5 0
3 years ago
Which statements describe how heat flows in foil?
LenKa [72]

Answer: c

Explanation:

8 0
3 years ago
Read 2 more answers
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes
frez [133]

Answer : The mass of MnCO_3 required are, 35 kg

Explanation :

First we have to calculate the mass of MnO_2.

The first step balanced chemical reaction is:

2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2

Molar mass of MnCO_3 = 115 g/mole

Molar mass of MnO_2 = 87 g/mole

Let the mass of MnCO_3 be, 'x' grams.

From the balanced reaction, we conclude that

As, (2\times 115)g of MnCO_3 react to give (2\times 87)g of MnO_2

So, xg of MnCO_3 react to give \frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg of MnO_2

And as we are given that the yield produced from the first step is, 65 % that means,

60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg

The mass of MnO_2 obtained = 0.4542x g

Now we have to calculate the mass of Mn.

The second step balanced chemical reaction is:

3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3

Molar mass of MnO_2 = 87 g/mole

Molar mass of Mn = 55 g/mole

From the balanced reaction, we conclude that

As, (3\times 87)g of MnO_2 react to give (3\times 55)g of Mn

So, 0.4542xg of MnO_2 react to give \frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg of Mn

And as we are given that the yield produced from the second step is, 80 % that means,

80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg

The mass of Mn obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of MnCO_3 required are, 35 kg

4 0
3 years ago
Read 2 more answers
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