Write 151.5 cm to meters in the correct significant figures

True or false, The four units that must always be used when using the ideal gas are 44.0 liters

Bogdan [553]

Answer:

for volume only liters can be used

Explanation:

4 0

3 years ago

While playing soccer, Caria slides across one of the lines on the field. When she stands up, what base does she stop?

SVEN [57.7K]

Answer:

Canxi hydroxit

Explanation:

8 0

2 years ago

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What is the limiting reactant when 5.6 moles of aluminum react with 6.2 moles of water?

alexira [117]

The limiting reactant when 5.6 moles of aluminium react with 6.2 moles of water is

water( H2O)

<u><em>Explanation</em></u>

The balanced equation is as below

2 Al +3 H2O → Al2O3 +3 H2

The mole ratio of Al :Al2O3 is 2:1 therefore the moles of Al2O3

= 5.6 x1/2 = 2.8 moles

The mole ratio of H2O: Al2O3 is 3:1 therefore the moles of Al2O3 produced

= 6.2 x1/3= 2.067 moles

since H2O yield less amount of Al2O3 , H2O is the limiting reagent.

7 0

3 years ago

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How many grams of Fe2O3 are formed when 16.7 moles of Fe reacts completely with

lesantik [10]

Answer:

The answer to your question is 1) 1336 g of Fe₂O₃ 2) 1 mol of H₃PO₄

Explanation:

1)

Data

mass of Fe₂O₃ = ?

moles of Fe = 16.7

Balanced chemical reaction

4Fe + 3O₂ ⇒ 2Fe₂O₃

a.- Calculate the moles of Fe₂O₃

4 moles of Fe ----------------- 2 moles of Fe₂O₃

16.7 moles of Fe -------------- x

x = (16.7 x 2)/4

x = 33.4/4

x = 8.35 moles of Fe₂O₃

b.- Calculate the molar mass of Fe₂O₃

Fe₂O₃ = (2 x 56)+ (3 x 16) = 112 + 48 = 160 g

c.- Convert moles to mass

160 g of Fe₂O₃ ------------ 1 mol

x ------------ 8.35 moles of Fe₂O₃

x = (8.35 x 160)/1

x = 1336 g of Fe₂O₃

2)

moles of H₃PO₄ = ?

mass of P₄O₁₀ = 71 g

Balanced chemical reaction

P₄O₁₀ + 6H₂O ⇒ 4H₃PO₄

a.- Calculate the molar mass of P₄O₁₀

P₄O₁₀ = (31 x 4) + (16 x 10) = 124 + 160 = 284 g

b.- Convert the mass of P₄O₁₀ to moles

284 g ------------------- 1 mol

71 g -------------------- x

x = (71 x 1)/284

x = 0.25 moles of P₄O₁₀

c.- Calculate the moles of H₃PO₄

1 mol of P₄O₁₀ ---------------- 4 moles of H₃PO₄

0.25 moles ----------------- x

x = (0.25 x 4) / 1

x = 1 mol of H₃PO₄

3 0

3 years ago

How much of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

Ivan

Answer: 2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=1\\M_1=0.5M\\V_1=?mL\\n_2=1\\M_2=0.05M\\V_2=25.0mL$

Putting values in above equation, we get:

$1\times 0.5\times V_1=1\times 0.05\times 25.0\\\\V_1=2.5mL$

Thus 2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

6 0

3 years ago