A train took 18 hours to finish a 1530 miles journey. What was

How many calories is 52 J of energy?

just olya [345]

Answer:

1J = 0.238846cal

52J = 12.42cal

5 0

3 years ago

If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make

vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

2H₂ + O₂ -----------> 2H₂O

Now look at the reaction for mole ratio

2H₂ + O₂ -----------> 2H₂O

2 mole 2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

2H₂ + O₂ -----------> 2H₂O

2 mole (2 g/mol) 2 mole (18 g/mol)

4 g 36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

4 g of H₂ ≅ 36 g of H₂O

30 g of H₂ ≅ X of H₂O

Do cross multiplication

X of H₂O = 30 g x 36 g / 4 g

X of H₂O = 270 g

Now convert grams of H₂O into moles

No. of moles = mass in grams/molar mass

Put values in above formula

No. of moles = 270 g / 18 (g/mol)

No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

5 0

3 years ago

A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

5 0

3 years ago

Predict the following product(s) of the following reaction? F2 + KCI →

slega [8]

Answer:

${ \rm{F _{2(g)} + 2KCl _{(aq)} → { \bold{2KF _{(g)} + Cl_{2(g)} }}}}$

Explanation:

» The prediction is 98% correct because single displacement reaction type is highly possible.

This is because Fluorine has is more electronegative than Chlorine in Potassium Chloride. So, it strongly displaces Chlorine from the solution hence forming Chlorine gas.

» The 2% of wrong prediction maybe because of wrong reactant measurements following mole concept chemistry.

If you are asked the observation,

Observation » A green yellowish gas is formed.

This gas is Chlorine gas (Cl2)

5 0

2 years ago

A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the

sergeinik [125]

The formula for molality---> m = moles solute/ Kg of solvent

the solute here is NH₃ because it's the one with less amount. which makes water the solvent.

1) let's convert the grams of NH₃ to moles using the molar mass

molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol

15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃

2) let's convert the grams of water into kilograms (just divide by 1000)

250.0 g= 0.2500 kg

3) let's plug in the values into the molality formula

molality= mol/ Kg---> 0.881 mol/ 0.2500 kg= 3.52 m

6 0

3 years ago