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2 years ago

Write down the word equation for the production of the following salts. Remember to include ALL reactants and ALL products.

Chemistry

1 answer:

irga5000 [103]2 years ago

7 0

Answer:

The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).

Explanation:

The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.

The substances generated by the reaction are called products, and their formulas are placed on the right sight of the equation.

Plus signs (+) separate individual reactant and product formulas, and an arrow (⟶) separates the reactant and product (left and right) sides of the equation.

The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

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What is the mass number of 207Pb ? A: 289 B: 207 C: 125 D: 82

Sergio [31]

207 is the mass number. 82 would be the atomic number

3 0

3 years ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0

3 years ago

What did Dalton's atomic theory contribute to science?

Serga [27]

Answer:

Explanation:

Pasted below is his 5 theory's and all of them are the building blocks of chemistry today.

1. Matter is made up of atoms that are indivisible and indestructible.

2. All atoms of an element are identical.

3. Atoms of different elements have different weights and different chemical properties.

4. Atoms of different elements combine in simple whole numbers to form compounds.

5. Atoms cannot be created or destroyed. When a compound decomposes, the atoms are recovered unchanged.

4 0

3 years ago

Calculate the volume of a 1.25 M solution of HCN made from 31 grams of HCN

fenix001 [56]

Answer:

V HCNsln = 0.9176 L

Explanation:

V HCNsln = ?

∴ m HCN = 31 g

∴ C HCNsln = 1.25 mol/L

∴ molar mass HCN = 27.0253 g/mol

⇒ V HCNsln = (31 g)*(mol/27.0253 g)*(L/1.25 mol) = 0.9176 Lsln

5 0

3 years ago

A chemical reaction is shown below:

BabaBlast [244]

Answer:

Mass = 8.46 g

Explanation:

Given data:

Mass of water produced = ?

Mass of glucose = 20 g

Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

C₆H₁₂O₆ : H₂O

1 : 6

0.11 : 6/1×0.11 = 0.66

O₂ : H₂O

6 : 6

0.47 : 0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol ×18 g/mol

Mass = 8.46 g

8 0

3 years ago