All categories

3 years ago

Balance equation for. _Mg + _H3(PO4) --_Mg3(PO4)2+ _H2

Chemistry

1 answer:

Katena32 [7]3 years ago

3 0

3Mg + 2H3(PO4) —> Mg3(PO4)2 + 3H2

I think

You might be interested in

In the reaction HCl + KOH KCL + H2O, the spectator ions are

zlopas [31]

KCl : K+ as cation and Cl- as anion

6 0

3 years ago

True or False

lozanna [386]

1. False (gravity)
2. True
3. False
4.True
5. False

Hopefully this helps :)

5 0

3 years ago

A 20g sample of iron at a temperature of 120oC is placed into a container of water. There are 300 milliliters of water in the co

galben [10]

Answer:

30.63 °C will be the final temperature of the water.

Explanation:

Heat lost by iron will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of iron = $m_1=20 g$

Specific heat capacity of iron = $c_1=0.444 J/g^oC$

Initial temperature of the iron = $T_1=120^oC$

Final temperature = $T_2$ =T

$Q_1=m_1c_1\times (T-T_1)$

Volume of water = 300 ml

Density of water = 1 g/mL

Mass of water= $m_2=300 mL\times 1 g/mL = 300 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=30^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

we get, T = 30.63°C

30.63 °C will be the final temperature of the water.

4 0

3 years ago

An area with an average annual temperature of more than 64°F and greater than 59 inches of annual rainfall would be an example o

aalyn [17]

Answer:

Based on the temperature and rainfall, it is the tropical

5 0

3 years ago

Read 2 more answers

A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a v

miskamm [114]

Answer:

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution $C_1=?$

Volume of the sport drink before dilution $V_1=5.4 mL$

Concentration of red dye in sport drink after dilution $C_2=18.1 ppm$

Volume of the sport drink after dilution $V_2=25.0 mL$

$C_1V_1=C_2V_2$ ( dilution )

$C_1=\frac{C_2V_2}{V_1}=\frac{18.1 ppm\times 25.0 mL}{5.4 mL}$

$C_1=83.80 ppm\approx 84.0ppm$

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

6 0

3 years ago