All categories

3 years ago

Dance ideas for solids: Dance ideas for liquids: Dance ideas for gases:

2 answers:

8 0

Maybe for solid u could do just standing still, for liquid you could do waving ur arms by ur side, and for gases you can wave your arms and legs around. I did the same thing last yr and that’s what I wrote, gl

hjlf3 years ago

7 0

Ok this is going to be really weird but in science I was told that the particles move slow and they’re closer together if they’re a solid, they move faster and more spaced apart for liquids and they move really really really fast and are really spaced out for gases. Hope this helps!

You might be interested in

If the formula for a compound is represented by X3Y2 and the charge on the Y ion is -3, what is the charge in the X ion?

artcher [175]

I cannot fully remember but the charge of a compound always has to be the same. However, I believe that the sign of the Y2 means that it is positive 2.

so in my opinion it is in between b and c. use your best judgment.

3 0

3 years ago

Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti

djverab [1.8K]

<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

= 80,000 g

Therefore;

$Moles of water = \frac{80,000g}{18.02 g/mol}$

= 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O available = 65.0 kg or 65,000 g

Molar mass Li₂O = 29.88 g/mol

Moles of Li₂O = 65,000 g ÷ 29.88 g/mol

= 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
This means, Li₂O is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles

= 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

= 4.0815 × 10⁴ g or

= 40.815 kg

Thus, the mass of excessive water is 40.815 kg

7 0

3 years ago

Compare and contrast chemical change and physical change

ki77a [65]

Well physical would be if you have Clay and you molded into a new shape or if you put butter on your toes and it melts or water evaporating from the surface of the ocean chemical changes would be milk going sour jewellery tarnishing which means turning into a different color or rust bread putting it in the oven and turning it into toes or rust forming on the nail that is left outside

3 0

4 years ago

Given a fixed amount of gas held at constant pressure, calculate the volume (in L) it would occupy if a 3.50 L sample were coole

Over [174]

Answer:

V₂ = 2.91 L

Explanation:

Given data:

Initial volume = 3.50 L

Initial temperature = 90.0°C (90+273 = 363 K)

Final temperature = 30.0 °C ( 30 +273 = 303 K)

Final volume = ?

Solution:

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

V₁/T₁ = V₂/T₂

3.50 L / 363 K) = V₂ / 303 K)

V₂ = 0.0096 L/K × 303 K

V₂ = 2.91 L

6 0

3 years ago

: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be

Jet001 [13]

Answer: New concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.

Explanation:

The given data is as follows.

Moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 mol

Moles of $H_{2}C_{6}H_{5}O_{7}$ = 0.08 mol

Therefore, moles of $OH^{-}$ added are as follows.

Moles of $OH^{-}$ = $0.125 \times \frac{25}{1000}$

= 0.003125 mol

Now, new moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 + 0.003125

= 0.063125

Therefore, new concentration of $HC_{6}H_{5}O^{2-}_{7}$ will be calculated as follows.

Concentration = $\frac{0.063125}{0.275}$

= 0.23 M

Thus, we can conclude that new concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.

4 0

3 years ago