All categories

3 years ago

How many hydrogen atoms are in 8.80 mol of ammonium sulfide

1 answer:

vladimir1956 [14]3 years ago

6 0

Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide. If the amount of ammonium sulfide is 8.9 moles, then the number of hydrogen atoms should be: 8/1 * 8.9 mol= 71.2 moles

You might be interested in

Uranium-238 decays very slowly.its half-life is 4.47 billion years.

coldgirl [10]

13.4 billion years is 3 times of the half-life, 4.47 billion years. So the Uranium-238 will go through three times of half decay. So the remain percentage will be 50%*50%*50%=12.5%.

6 0

2 years ago

Read 2 more answers

when methane reacts with oxygen, the products are carbon dioxide and water. How many grams of carbon dilxide are formed if 30 g

Wittaler [7]

Answer:

CH4+2O2→CO2+2H2O

Moles of oxygen=3232=1

Moles of carbon dioxide=4422=0.5

Moles of water=1818=1

1 mole of methane reacts with 2 moles of oxygen to give 1 mole of CO2 and 2 moles of water so 0.5 mole of methane will be required which is equal to 0.5×16=8g.

Explanation:

4 0

2 years ago

The atmosphere is a(n) _____. <br> 1. element <br> 2. compound <br> 3. mixture

tensa zangetsu [6.8K]

The atmosphere is a mixture of gasses.

5 0

2 years ago

Read 2 more answers

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

2 years ago

The two methods in the lab for measuring the volume of a geometric solid object are weighing it on the balance and measuring it

Yakvenalex [24]

Answer:

Measuring with a ruler and using final volume minus initial volume

Explanation:

You can measure the volume of a geometric object by measuring its sides with a ruler and calculating the volume according to the corresponding formula for each object. For example, for a rectangular prism it would be

$volume=length*width*height$

You can also measure the volume of an object by measuring how much water it displaces. To do this you have to fill a measuring cylinder with enough water for the object to be completely submerged and take note of the volume. Then, add the object and note again the volume of the water+object. The difference between both is the volume of the object.

$Volume of the object= volume of water and object - volume of water$

The advantage of the second method is that it can be used for objects with irregular shapes as long as they do not float.

8 0

3 years ago