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3 years ago

Write the net ionic equation for the precipitation of silver phosphate from aqueous solution:________

Chemistry

1 answer:

SVEN [57.7K]3 years ago

5 0

Answer: 3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)

Explanation:

The question requires us to write down the net ionic equation for the precipitation of silver phosphate from aqueous solution:________

Note that we we told to:

(Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)

Answer:

From the solubility rule,

Phosphate (PO₄³⁻) and Carbonates are insoluble except for NH₄⁺ and Group1A compounds.

According to the solubility rule, the precipitation of silver phosphates and that of the net ionic equations is given thus;

3AgNO³ (aq) + K₃PO₄ (aq) → Ag₃PO₄ (s) + 3KNO₃ (aq)

while the Net ionic equation is given;

3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)

cheers i hope this helped !!

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Use the dichotomous key below to identify the cell type.

xenn [34]

The dichotomous key is not given here, however, the correct dichotomous key is as follows:

step- 1a cell has a nucleus and organelles - go to step 2

1b cell has no nucleus or organelles - bacterial cell

step- 2a cell has a cell wall - go to step 3

2b cell has no cell wall - animal cell

step- 3a cell has chloroplasts - plant cell

3b cell has no chloroplasts - fungal cell

Using this dichotomous key the cell can be identified as - Fungal cell.

A dichotomous key is an essential scientific tool that is used to find out the various kinds of organisms on their observable traits. Dichotomous keys have a series of statements with two different choices in each step that will take the student or user to the correct identification.

Identification of the given cell:

1. Follow step 1.

If cells have a nucleus and organelles - yes then go to step 2

2. follow step 2.

2a. if cell has a cell wall - yes go to step 3

3. follow step 3.

3a. if cell has chloroplasts - no then fungal cell

Fungal cells have a nucleus and cell organelles enclosed with a cell wall around the plasma membrane but do not have chloroplast.

Thus, Using this dichotomous key the cell can be identified as - Fungal cell.

Learn more about fungal cells:

brainly.com/question/14470940

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3 years ago

Which of these statements best describes the relationship between matter, pure substances, and mixtures?

borishaifa [10]

The answer you are looking for is the second option, "<span>Matter can be divided into two groups: pure substances and mixtures."

A pure substance can describe a pure element or chemically bonded elements. A pure substance cannot be broken down into further substances. An example of these would be Fe (iron) or O2 (oxygen).

A mixture is a structure made from physically combining different non-bonded substances. An example of a mixture is blood.

Both pure substances and mixtures are types of matter.</span>

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4 years ago

A gas at 20.0°C and 52.1 mL is heated to 93.5°C. What is the new volume?

lawyer [7]

$\frac{ V_{1} }{ T_{1} }$ = $\frac{ V_{2} }{ T_{2} }$
$V_{1}$ = 0.0521 L
$T_{1}$ = 293.15 K
$T_{2}$ = 366.65 K

Solve for $V_{2}$

$\frac{0.0521 L}{293.15 K}$ = $\frac { V_{2} }{366.65 K}$

$V_{2}$ = 0.06516 L or 65.2 mL

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3 years ago

How did the discoveries of scandium (sc) and germanium (ge) affect mendeleevâs work?

lukranit [14]

they showed mandeleeves predictions were correct

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4 years ago

Read 2 more answers

Determine the mass of grams of led(II) sulfate that will dissolve in 2.50 x 10^2 ml water. Ksp= 1.8 x 10^-8

ki77a [65]

Answer:

9.86*10^(-3) g

Explanation:

PbSO4 ----> Pb^(2+) + SO4^(2-)

s s

Ksp = s²

s =√Ksp = √(1.8*10^-8) = 1.3*10^(-4) mol/L

The molar solubility PbSO4 = 1.3*10^(-4) mol/L.

2.50 *10^2 mL *1L/10³mL =0.250L

1.3*10^(-4)mol/L *0.250L*303.3 g/mol = 9.86*10^(-3) g

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3 years ago