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2 years ago

KOH+H 3 PO 4 K3PO4+ H₂O How to solve and answer to this question

Chemistry

1 answer:

sergij07 [2.7K]2 years ago

4 0

Answer:

hmmmm

Explanation: Balance the reaction of KOH + H3PO4 = K3PO4 + H2O using this chemical equation balancer!

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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

2 years ago

A 3000 MWt fast reactor has a 42% efficiency. This reactor operates for 23 months followed by a 1 month down time for refueling

n200080 [17]

Answer:

capacity factor = 0.952

Availability factor = 0.958

Explanation:

1) capacity factor

capacity factor = actual power output / maximum power output

= (actual power output)/(efficiency * rated power output)

$= \frac{1200}{\frac{42}{100}*3000}$

= 0.952

2) Availability factor

Availability factor = Actual operation time period/ total time period

= 23/24 = 0.958

8 0

2 years ago

During the combustion of 2.00 g of coal, the temperature of 500 g of water inside the calorimeter increased from 25.0°c to 43.7°

pantera1 [17]

Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, difference at temperatures.
c(water) = 4,18 J/g·°C, specific heat of water
Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.

6 0

3 years ago

The most concentrated solution is

aev [14]

The most concentrated solution is b

6 0

3 years ago

Read 2 more answers

What effect does mass have on an objects density?

xz_007 [3.2K]

The density of an object or quantity of matter is its mass divided by its volume.

6 0

3 years ago