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3 years ago

What subatomic particle acts like a mini-magnet?

1 answer:

5 0

The subatomic particles that acts like a mini-magnet is electron. Electrons are negatively charged sub atomic particles in an atom. The electron spin is a property of an electron that makes it behave like it's spinning; a spinning electron produces a magnetic field that makes it behave like a tiny magnet in an atom.

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A 0.89 m aqueous solution of an ionic compound with the formula mx has a freezing point of -3.0 ∘c . van't hoff factor?

BigorU [14]

Answer is: V<span>an't Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).

i = 1,81.

5 0

3 years ago

When baking soda and vinegar are mixed together bubbles form and the vinegar and baking soda are changed into new chemicals. Thi

nadya68 [22]

Answer:

A. Chemical change

Explanation:

When vinegar and baking soda are first mixed together, hydrogen ions in the vinegar react with the sodium and bicarbonate ions in the baking soda. The result of this initial reaction is two new chemicals: carbonic acid and sodium acetate.

3 0

3 years ago

1. What affects a material's resistance?

Ghella [55]

Answer:

A. Thickness and temperature

Explanation:

3 0

3 years ago

A ball is thrown horizontally from a 19 m -high building with a speed of 2.0 m/s . How far from the base of the building does th

bazaltina [42]

Using the formula: ΔY = V₀y * t + (1/2) * ay * t²

Solve for time and get: 1.968s

Then use: v = d/t in the x-direction and get: d = 3.936

3 0

3 years ago

Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re

lapo4ka [179]

Answer:

$K_{system} = \frac{k}{10}$

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})$

Here, n = 10

Spring constant of each spring = k

Thus,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})$

$\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})$

$\frac{1}{K_{system}} = \frac{10}{k}$

$K_{system} = \frac{k}{10}$

7 0

3 years ago

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