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2 years ago

I need help with question 2 I will brainliest you if I’m able to

Chemistry

2 answers:

erik [133]2 years ago

4 0

Answer:

Blank 1: Microscope

Blank 2: Cells

vfiekz [6]2 years ago

3 0

Answer: (answers are in bold)

Because of the invention of the microscope, we know that all organisms are made of small living creatures called cells.

Explanation:

Well the statement explains itself; I'll explain the invention of the microscope.

No one really knows who invented the simple microscope, but the compound microscope was invented c. 1590 by Dutch lens-maker Zacharias Janssen. He used it to magnify object to about 10x their original size. Anton von Leeuwenhoek made even better ones; those had up to 275x magnification.

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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.

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0

2 years ago

Pls give help me out>!

iris [78.8K]

Answer:

3 because it the element that combined the form

7 0

2 years ago

How many moles are 3.20 x102 formula units of calcium iodide?

DedPeter [7]

The number of moles in 3.20 x 10² formula units of calcium iodide is 0.053 moles.

<h3>How to calculate number of moles?</h3>

The number of moles in the formula units of a substance is calculated by dividing the formula unit by Avogadro's number.

According to this question, 3.20 x 10² formula units are in calcium iodide. The number of moles is as follows:

no of moles = 3.20 x 10²² ÷ 6.02 × 10²³

no of moles = 0.53 × 10-¹

no of moles = 0.053 moles

Therefore, the number of moles in 3.20 x 10² formula units of calcium iodide is 0.053 moles.

Learn more about number of moles at: brainly.com/question/12513822

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3 0

2 years ago

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How many moles are in 25.2g of KMnO4?

Zolol [24]

When finding the moles in a compound you have to know the grams. In this case, 25.2 grams are given for KMnO4. To find the moles you would divide the amount of grams by the molar mass of KMnO4. The molar mass of KmnO4 is 158.034. You you would now divide 25.2 by 158.034 which is 0.15946 moles. Depending on how many decimal places the questions asks for is dependent on you. I just went with 5 significant figures.

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2 years ago

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Which combination of factors is most suitable for increasing the electrical conductivity of metals

Lemur [1.5K]

More valence electrons and larger atomic radius are facts most suitable for increasing the electrical conductivity of metals.

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3 years ago

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