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4 years ago

Melted candle wax has no fixed shape, but it does have a fixed volume. So, melted candle wax is a

2 answers:

7 0

It is a liquid because when you have a liquid, there is no definite shape. Therefore, this would be the answer because it takes the shape of its container.

Final answer: a. Liquid

xz_007 [3.2K]4 years ago

5 0

The answer is A.Liquid

You might be interested in

Water has a density of 1 g/cc. If you have 1 cc of water, how much mass does it contain?

rewona [7]

Water has a density of 1 g/cc. If you have 1 cc of water, then the mass it contains would be 1 g. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance. Hope this answers the question. Have a nice day.

6 0

4 years ago

ethanol is a common laboratory solvent and has a density of 0.789 g/mL. what is the mass, in grams, of 151 mL of ethanol

natta225 [31]

V = 151 mL = 151 cm³
d = 0,789 g/mL = 0,789 g/cm³
--------------------------------------

d = m/V
m = d×V
m = 0,789×151
m = 119,139g

3 0

4 years ago

Read 2 more answers

Need help ASAP please!

bearhunter [10]

Answer:

13 mol NO

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of O₂ to NO is 5:4.

Step 3: Calculate the number of moles of O₂ needed to produce 16 moles of NO

We will use the previously established molar ratio.

16 mol O₂ × 4 mol NO/5 mol O₂ = 13 mol NO

3 0

3 years ago

A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of

jek_recluse [69]

Answer: The volume when the pressure and temperature has changed is $1.6\times 10^2mL$

Explanation:

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

Let us assume:

$P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K$

Putting values in above equation, we get:

$\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL$

Hence, the volume when the pressure and temperature has changed is $1.6\times 10^2mL$

5 0

3 years ago

Katarina [22]

The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:

From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.

So, ∑Fx = 0 and ∑Fy = 0

Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,

∑Fx = -Fapp + Ffric = 0

-Fapp + Ffric = 0

Fapp = Ffric

Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,

∑Fy = Fnorm + (-Fgrav) = 0

Fnorm - Fgrav = 0

Fnorm = Fgrav

So, we see that the size of Fapp equals size of Ffric and the size of Fnorm equals the size of Fgrav.

So, the correct option is C

The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.

Learn more about equilibrium of forces here:

brainly.com/question/12980489

5 0

2 years ago