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3 years ago

Which of these substances are most likely crystalline solids? Select all that apply.

Chemistry

2 answers:

Lostsunrise [7]3 years ago

8 0

Answer:

you did not show the options

Explanation:

Svet_ta [14]3 years ago

5 0

I am unsure if this is the correct question, but I think these are the answer choices that are part of this question:

O diamond

O sugar

O rubber

O salt (NaCl)

O wood

O ice

O flour

The answers to this question are diamond, sugar, salt (NaCl), and ice (1st, 2nd, 4th, and 6th options).

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Water often has dissolved gases and other chemicals in it. When this water encounters rocks, the minerals in the rocks can be ch

babunello [35]

Erosion is the best anwser for that

5 0

3 years ago

A solution contains one or more of the following ions: Ag+, Ca2+, and Co2+. Lithium bromide is added to the solution and no prec

olchik [2.2K]

<h3>Answer:</h3>

#1. Ca²⁺

# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

<h3>Explanation:</h3>

The question above concerns solubility of salts or ions in water.

The solution given contains Ag+, Ca2+, and Co2+ ions.

In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.

The net ionic equation for the reaction is;

Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.

In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.

The net ionic equation for the reaction is;

3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.

6 0

3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

6 0

3 years ago

What is the enthalpy change (in kj) of a chemical reaction that raises the temperature of 250.0 ml of solution having a density

Pepsi [2]

<span>250 ml * 1.25 g/ml * 3.74 j/g-K * 9.2 K = 10.752 kJ Pretty much, all you need to do here is multiply all of these out to get your final answer. Not all questions are this easy, but this is certainly one of them.</span>

3 0

3 years ago

Read 2 more answers

Diethyl ether is produced from ethanol according to the following equation:

juin [17]

The balanced chemical equation is given as:

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.

Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3

% yield = .775 = actual yield / 264.66

actual yield = 205.11 g CH3CH2OCH2CH3

6 0

3 years ago