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3 years ago

How are the particles in solid ice similar to the particles in liquid water, and water vapor? How are they different?

Chemistry

1 answer:

ki77a [65]3 years ago

5 0

Answer:

The particles in most solids are closely packed together. Even though the particles are locked into place and cannot move or slide past each other, they still vibrate a tiny bit. ... However, ice is different from most solids: its molecules are less densely packed than in liquid water. This is why ice floats.

Explanation:

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An isotope of hydrogen commonly referred to as heavy water is

muminat

Answer:

Heavy water is a compound that is made up of oxygen and deuterium, a heavier isotope of hydrogen which is denoted by ‘ 2 H’ or ‘D’. Heavy water is also called deuterium oxide and is denoted by the chemical formula D 2 O.

Density: 1.107 g/mL

Dipole moment: 1.87 D

Heavy Water (Deuterium Oxide): D2O

Molecular Mass: 20.02 grams/mole

Explanation:

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3 years ago

Is lysomes a plant or animals cell

sergeinik [125]

Answer:

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Explanation:

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4 years ago

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Which of the following describes the stage that occurs after evaporation in the water cycle?

saw5 [17]

C. Condensation is the right answer

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4 years ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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2 years ago

A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure has its volume increased

allochka39001 [22]

Answer:

2.5 atm

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15k

P2 = 2.0atm

P1 = ?

From general gas equation,

(P1 × V1) / T1 = (P2 × V2) / T2

P2 × V2 × T1 = P1 × V1 × T2

P1 = (P2 × V2 × T1) / (V1 × T2)

P1 = (2.0 × 34 × 318.15) / (28 × 308.15)

P1 = 21634.5 / 8628.2

P1 = 2.5 atm

The initial pressure of the gas is 2.5atm

5 0

3 years ago