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4 years ago

True or false a contour line indicates that all points on that line have equal elevation

1 answer:

3 0

True. Each contour line on a topographic map represent a certain elevation; the closer they are, the steeper the surface is. All points on a contour line are the same elevation.

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A balanced chemical reaction obeys the law of...

tia_tia [17]

Its c. Not a 100% sure

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3 years ago

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What is difference between vapourization and boiling point?

Pepsi [2]

Answer:

Vaporization is basically just evaporation which does not require nearly as much heat and happens naturally however when it comes to boiling point this is the point where water has been heated enough to start bubbling and changing state slowly.

Explanation:

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3 years ago

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DUE TODAY EXTREMELY IMPORTANT PLEASE PLEASE PLEASE HELP ME ASAPPPPPPPP CHEMISTRY!!!!!!!!!!!!!!!!!!! I WOULD APPRECIATE IT ALOT T

musickatia [10]

Answer:

Here's what I get

Explanation:

3. Molar concentration by formula.

$\begin{array}{rcl}M_{\text{a}}V_{a} & = & M_{\text{b}}V_{b}\\M_{a} \times \text{0.025 00 L} & = & \text{0.3840 mol/L} \times \text{0.034 52 L}\\0.025 00M_{a}\text{ L} & = & \text{0.013 26 mol}\\M_{a}&= &\dfrac{\text{0.013 26 mol}}{\text{0.025 00 L}}\\\\& = &\textbf{0.5302 mol/L}\end{array}$

(i) Comparison of molar concentrations

The formula gives a calculated value of 0.5302 mol·L⁻¹.

Dimensional analysis gives a calculated value of 0.1767 mol·L⁻¹.

The first value is three times the second.

It is wrong because the formula assumes that the acid supplies just enough moles of H⁺ to neutralize the OH⁻ from the NaOH.

Instead, I mol of H₃PO₄ provides 3 mol of H⁺, so your calculated concentration is three times the true value.

(ii) When is the formula acceptable?

The formula is acceptable only when the molar ratio of acid to base is 1:1.

Examples are

HCl + NaOH ⟶ NaCl + H₂O

H₂SO₄ + Ca(OH)₂ ⟶ CaSO₄ + 2H₂O

H₃PO₄ + Al(OH)₃ ⟶ AlPO₄ + 3H₂O

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3 years ago

What happens to the cool and hot zones of a nonluminous flame when the air control valve is closed?

lesya [120]

When the cool and hot zones of a non luminous flame’s air control valve is closed, the gas in which will be in placed as first and as it encounters air as it was in the mouth of the burner, there will be a presence of incomplete mixing, causing a combustion which would be incomplete and will also lead to a process that can’t be controlled, resulting to flame that is cool.

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4 years ago

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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