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Gwar [14]
3 years ago
5

Two children on a seesaw are able to balance perfectly while on Earth.would they still be balanced if you saw was brought to the

Moon?
A no,they would not be balanced because their masses would change

b yes,they would still be balanced because their weights would not change

c yes,they would still be bounced because their masses would not change

d no,they will not be bounced because their weights could change​
Chemistry
1 answer:
Setler [38]3 years ago
3 0

Answer:

C

Explanation:

Because on the moon you are basically weightless and your mass would stay the same

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Glass of orange juice contain 0.90 dL of juice. How many milliliters of orange juice is that?
dlinn [17]
.90 dL is 90 mL because 1 dL is 100 mL
6 0
3 years ago
Can I find a tutor to help me with this question?
Eddi Din [679]

INFORMATION:

We have the following statements

And we must complete them

STEP BY STEP EXPLANATION:

To complete the statements, we need to classify matter according to its state:

- Solid:

there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume.

- Liquid:

That describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.

- Gas:

That describes the gas state, which we will consider in more detail elsewhere. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either.

Finally, we know that:

- A solid has a definite volume and has a definite shape

- A liquid has a definite volume and has not a definite shape

- A gas has not a definite volume and has not a definite shape

ANSWER:

- A solid has a definite volume and has a definite shape

- A liquid has a definite volume and has not a definite shape

- A gas has not a definite volume and has not a definite shape

8 0
1 year ago
4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.
Bas_tet [7]
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

7 0
3 years ago
If .758 moles of gas occupy a volume of 80.6L, how many moles will occupy a volume of 270.9L?
egoroff_w [7]

Answer:

n₂ = 2.55 mol

Explanation:

Given data:

Initial number of moles = 0.758 mol

Initial volume = 80.6 L

Final volume = 270.9 L

Final number of moles = ?

Solution:

Formula:

V₁/n₁ = V₂/n₂

V₁ = Initial volume

n₁ = initial number of moles

V₂ = Final volume

n₂ =  Final number of moles

now we will put the values in formula.

V₁/n₁ = V₂/n₂

80.6 L / 0.758 mol = 270.9 L/ n₂

n₂ = 270.9 L× 0.758 mol / 80.6 L

n₂ =  205.34 L.mol /80.6 L

n₂ = 2.55 mol

4 0
3 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0
3 years ago
Read 2 more answers
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