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3 years ago

Ensure you write the states of the reactants and products.

Chemistry

2 answers:

slamgirl [31]3 years ago

6 0

<h3>Answer:</h3>

Phosphoric acid reacts with magnesium hydroxide to produce magnesium phosphate and water via the following reaction:

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

(solid) (solid) (solid) (liquid)

<h3>Explaination:</h3>

This is a typical neutralization reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat,

ΔH < 0 , and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.

Doing so we obtain :

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

(solid) (solid) (solid) (liquid)

<h3>hope it helps :)</h3>

TiliK225 [7]3 years ago

4 0

H3PO4 (aq) + Mg(OH)2 (s) -> Mg3(PO4)2 (s) + H2O (l)

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What is the density of a liquid that has a mass of 27g and a volume of 30mL?

NNADVOKAT [17]

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3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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