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stepladder [879]
3 years ago
14

In the desert, wind picks up and carries fine particles of sand and dirt. As 5 points

Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

C

It’s wrong I need points

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Which of the following salts will be less soluble in 0.10 M NaCl than it is in pure water?
MariettaO [177]

The less soluble salt : PbCl₂

<h3>Further explanation</h3>

Given

0.1 M NaCl

Required

The less soluble salt

Solution

If we see from the answer option, the salt that is more difficult to dissolve in NaCl is PbCl₂ because it has the same ion (Cl)

When PbCl₂ is dissolved in water, ionization will occur

PbCl₂ ⇒ Pb²⁺+ 2Cl⁻

So, when dissolved in NaCl, NaCl itself will be ionized

NaCl ⇒ Na⁺ + Cl⁻

Based on the principle of equilibrium, the addition of an ion (one of the ions is enlarged), the reaction will shift towards the ion that was not added. In addition to this Cl ion, the reaction will shift to the left so that the solubility of PbCl₂ will decrease (the reaction to the right decreases)

8 0
3 years ago
Melted rock can ooze out from below earths surface through a crack in the crust called a(n)?
Katen [24]
Melted rock can ooze out from below the earth's surface through a crack called a fault.
5 0
3 years ago
A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon
svetlana [45]

Answer:

\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}

Explanation:

1. Miles travelled in an average month  

\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}

2. Using a gasoline powered vehicle  

(a) Moles of heptane used  

n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}

(b) Equation for combustion  

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O  

(c) Moles of CO₂ formed  

n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}

(d) Volume of CO₂ formed  

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.  

V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}

3. Using an electric vehicle  

(a) Theoretical energy used  

\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}

(b) Actual energy used  

The power station is only 85 % efficient.  

\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

The heat of combustion of methane is -802.3 kJ·mol⁻¹  

V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}

4. Comparison  

\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}

6 0
3 years ago
27/13Al+4/2He &gt;&gt;&gt;&gt; 30/15P+
Veseljchak [2.6K]

Answer

Explanation:

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

5 0
3 years ago
What identifies an ion
bezimeni [28]

Answer:

Identifying whether or not an element is an ion is a very simple process. Identify the charge of the element. ... The number of electrons is equal to the atomic number minus the charge of the atom. Refer to an element with either a positive or negative charge as an ion.

7 0
3 years ago
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