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3 years ago

A pure substance is matter that consists of matter with a composition that

Chemistry

2 answers:

sashaice [31]3 years ago

7 0

Is constant (matter that has a composition that is the same everywhere)

tigry1 [53]3 years ago

6 0

Answer:

A Pure substances is a substances that are made of only one type of atom or molecule and it possesses the following physical properties

1. well-defined melting

2. Well-defined boiling points.

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Give free 18 points again just answer yes or no is einstein a scientist? yes or no

BARSIC [14]

Answer:

yes

Explanation:

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2 years ago

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Which statement would support a merit of the Bronsted-Lowry base theory has over the Arrhenius base theory?

butalik [34]

Answer:

Explanation:

Bronsted Base is an H+ acceptor

No good answer Bronstead base does not accept hydroxide or electrons

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2 years ago

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A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and 32 C.

lisabon 2012 [21]

Answer:

1255.4L

Explanation:

Given parameters:

P₁ = 928kpa

T₁ = 129°C

V₁ = 569L

P₂ = 319kpa

T₂ = 32°C

Unknown:

V₂ = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

take the units to the appropriate ones;

kpa to atm, °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁ = 928kpa gives 9.16atm

T₂ = 32°C gives 273 + 32 = 305K

T₁ = 129°C gives 129 + 273 = 402K

Input the values in the equation and solve for V₂;

$\frac{9.16 x 569}{402} = \frac{3.15 x V_{2} }{305}$

V₂ = 1255.4L

4 0

3 years ago

(IMAGE ATTACHED) Help plssss this is due in 3 hours

Natasha2012 [34]

Answer:

Acceleration:

Speed/Time

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Speed:

Distance/Time

Change in distance over a specific amount of time

Velocity:

Distance/Time

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2 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

2 years ago