Water is a solute. solvent.

Compared to winds on the earth the winds on Saturn are much stronger. <br><br> True or False?

Yanka [14]

This is most definitly true

3 0

2 years ago

Which equation best represents the net ionic equation for the reaction that occurs when aqueous solutions of potassium phosphate

laiz [17]

Answer:

2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)

Explanation:

In a net ionic equation you list <em>only the ions that are participating in the reaction. </em>

When potassium phosphate, K₃PO₄, reacts with iron (II) nitrate, Fe(NO₃)₂ producing iron (II) phosphate, Fe₃(PO₄)₂ that is an insoluble salt. The reaction is:

2K₃PO₄ + 3 Fe(NO₃)₂ → Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

The ionic equation is:

6K⁺ + 2PO₄³⁻ + 3Fe²⁺ + 6NO₃⁻→ Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

Subtracting the K⁺ and NO₃⁻ ions that are not participating in the reaction, the net ionic equation is:

8 0

3 years ago

CALCULATOR

suter [353]

Answer: The possible molecular formula will be $CO_2$

Explanation:

Mass of C= 27.3 g

Mass of O = 72.7 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{27.3g}{12g/mole}=2.275moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{72.7g}{16g/mole}=4.544moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{2.275}{2.275}=1$

For O = $\frac{4.544}{2.275}=2$

The ratio of C : O = 1: 2

Hence the empirical formula is $CO_2$

The possible molecular formula will be= $n\times CO_2$

5 0

2 years ago

An oily liquid with a cheesy, waxy odor like that of goats or other barnyard animals is caproic acid, a compound containing carb

Zinaida [17]

The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O

We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:

How to determine the mass of C

Mass of CO₂ = 9.78 g
Molar mass of CO₂ = 44 g/mol
Molar of C = 12 g/mol
Mass of C =?

Mass of C = (12 / 44) × 9.78

Mass of C = 2.67 g

How to determine the mass of H

Mass of H₂O = 20.99 g
Molar mass of H₂O = 18 g/mol
Molar of H = 2 × 1 = 2 g/mol
Mass of H =?

Mass of H = (2 / 18) × 4

Mass of H = 0.44 g

How to determine the mass of O

Mass of compound = 4.30 g
Mass of C = 2.67 g
Mass of H = 0.44 g
Mass of O =?

Mass of O = (mass of compound) – (mass of C + mass of H)

Mass of O = 4.30 – (2.67 + 0.44)

Mass of O = 1.19 g

<h3>How to determine the empirical formula </h3>

The empirical formula of the compound can be obtained as follow:

C = 2.67 g
H = 0.44 g
O = 1.19 g
Empirical formula =?

Divide by their molar mass

C = 2.67 / 12 = 0.2225

H = 0.44 / 1 = 0.44

O = 1.19 / 16 = 0.074

Divide by the smallest

C = 0.2225 / 0.074 = 3

H = 0.44 / 0.0744 = 6

O = 0.074 / 0.074 = 1

Thus, the empirical formula of the compound is C₃H₆O

Learn more about empirical formula:

brainly.com/question/9459553

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6 0

11 months ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

kaheart [24]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

6 0

2 years ago