True or false?<br><br> Freshwater can occur as a gas, liquid, or solid.

Tides describe the regular rising and falling of ocean water. Tides are caused by

user100 [1]

Tides are the rise and fall of the oceans. They are caused by the gravity, or pull, of the Moon and Sun. The Moon's gravity is the main force in causing tides. It makes the oceans bulge out toward it. Another bulge occurs on the opposite side, because Earth is being pulled toward the Moon and away from the water. The water on the side farthest away from the Moon is least affected by its gravity.

3 0

3 years ago

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Nitrogen forms more oxides than any other element. The percents by mass of in three different nitrogen oxides are (1) (II) and (

Tema [17]

Complete question;

Nitrogen forms more oxides than any other element. The percents by mass of N in three different nitrogen oxides are (|) 46.69%;(II) 36.85 %; (III) 25.94%. For each compound, determine (a) the simplest whole-number ratio of N to O, and (b) the number of grams of oxygen per 1.00 g of nitrogen.

Answer:

a. (i) The ratio is 1:1 , the formula = NO (ii)The ratio is 1 : 1.5 which is 2 : 3, the formula = N₂O₃ (iii) The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅

b. (i)number of grams of oxygen = 53.31/46.69 = 1.14 g

(ii)number of grams of oxygen = 63.15/36.8 = 1.71 g

(iii)number of grams of oxygen = 74.06/25.94 = 2.855 g

Explanation:

a.

(i) The percentage by mass of the nitrogen in Nitrogen oxide (i) is 46.69% which is taken as 46.69 grams . Since the other element is oxygen the mass of oxygen will be 100 - 46.69 = 53.31 grams.

The relative atomic mass of Nitrogen and oxygen is 14 amu and 16 amu respectively.

Therefore, to know the whole number ratio of N and O we find the number of moles.

number of moles of N = 46.69/14 = 3.335

number of moles of O = 53.31/16 = 3.332

The ratio is 1:1 , the formula = NO

(ii)

number of moles of N = 36.85/14 = 2.632

number of moles of O = 63.15/16 = 3.947

The ratio is 1 : 1.5 which is 2 : 3, the formula = N₂O₃

(iii)

number of moles of N = 25.94/14 = 1.85

number of moles of O = 74.06/16 = 4.63

The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅

b.

(i) 46.69 g of nitrogen = 53.31 g of oxygen

1 g of nitrogen = ? of Oxygen

number of grams of oxygen = 53.31/46.69 = 1.14 g

(ii)

Using similar method in b(i)

number of grams of oxygen = 63.15/36.8 = 1.71 g

(iii)

Using similar method in b(i)

number of grams of oxygen = 74.06/25.94 = 2.855 g

3 0

3 years ago

Where does the greenhouse effect occur?

attashe74 [19]

Answer:

the greenhouse effect occurs in the earths atmosphere

7 0

3 years ago

What does br mean on the periodic table?

sineoko [7]

BR means breaking bad on a periodic table

8 0

3 years ago

Freeze-drying is a process used to preserve food. If strawberries are to be freeze-dried, then they would be frozen to -80.00 °C

katrin2010 [14]

Answer:

1. 389 kJ; 2. 7.5 µg; 3. 6.25 days

Explanation:

1. Energy required

The water is converted directly from a solid to a gas (sublimation).

They don't give us the enthalpy of sublimation, but

$\Delta_{\text{sub}}H = \Delta_{\text{fus}}H + \Delta_{\text{vap}}H = 6.01 + 40.68 = 46.69 \text{ kJ}\cdot\text{mol}^{-1}$

The equation for the process is then

Mᵣ: 18.02

46.69 kJ + H₂O(s) ⟶ H₂O(g)

m/g: 150

(a) Moles of water

$\text{Moles} = \text{150 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{8.324 mol}$

(b) Heat removed

46.69 kJ will remove 1 mol of ice.

$\text{Heat removed} = \text{8.234 mol} \times \dfrac{\text{46.69 kJ}}{\text{1 mol}} = \textbf{389 kJ}\\\text{It takes $\large \boxed{\textbf{389 kJ}}$ to remove 150 g of ice}$

2. Mass of water vapour in the freezer

For this calculation, we can use the Ideal Gas Law — pV = nRT

(a) Moles of water

Data:

$p = 1.00 \times 10^{-3}\text{ torr } \times \dfrac{\text{1 atm}}{\text{760 torr}} = 1.316 \times 10^{-6}\text{ atm}$

V = 5 L

T = (-80 + 273.15) K = 193.15 K

Calculation:

$\begin{array}{rcl}pV & = & nRT\\1.316 \times 10^{-6}\text{ atm} \times \text{5 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{193.15 K }\\6.6 \times 10^{-6} & = & 15.85n\text{ mol}^{-1} \\n & = & \dfrac{6.6 \times 10^{-6}}{15.85\text{ mol}^{-1}}\\\\& = & 4.2 \times 10^{-7} \text{ mol}\\\end{array}$

(b) Mass of water

$\text{Mass} = 4.2 \times 10^{-7} \text{ mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = 7.5 \times 10^{-6}\text{ g} = 7.5 \, \mu \text{g}\\\\\text{At any given time, there are $\large \boxed{\textbf{7.5 $\mu$g}}$ of water vapour in the freezer.}$

3. Time for removal

You must remove 150 mL of water.

It takes 1 h to remove 1 mL of water.

$\text{Time} = \text{150 mL} \times \dfrac{\text{1 h}}{\text{1 mL}} = \text{150 h} = \text{6.25 days}$

5 0

4 years ago

True or false? Freshwater can occur as a gas, liquid, or solid.