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julia-pushkina [17]
1 year ago
12

Which number is equal to –906,060?

Chemistry
1 answer:
mamaluj [8]1 year ago
7 0
In absolute value it is equal to 906,060
In scientific notation it is equal to -9.06,060^5
You might be interested in
) Starting with 5.00 g barium chloride n hydrate yields 4.26 g of anhydrous barium chloride after heating. Determine the integer
RSB [31]

Answer:

2

Explanation:

Mass of water molecule = mass of hydrated salt - mass of anhydrous salt

Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.

Number of moles = mass / molarmass

Molar mass of water = 18.015g/mol

No. of moles of water = 0.74 / 18.015 = 0.0411 moles.

Mass of BaCl2 present =?

1 mole of BaCl2 = 208.23 g

X mole of BaCl2 = 4.26 g

X = (4.26 * 1) / 208.23

X = 0.020

0.020 moles is present in 4.26g of BaCl2

Mole ratio between water and BaCl2 =

0.0411 / 0.020 = 2

Therefore 2 molecules of water is present the hydrated salt.

4 0
2 years ago
How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge
Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:-  The balanced equation for the combustion of ethane is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})

= 126 L O_2

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

105.0LO_2(\frac{4LCO_2}{7L O_2})

= 60.0 L CO_2

Similarly, let's calculate the volume of water vapors formed:

105.0L O_2(\frac{6L H_2O}{7L O_2})

= 90.0 L H_2O

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

105.0LO_2(\frac{2LC_2H_6}{7LO_2})

= 30.0 L C_2H_6

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

5 0
1 year ago
Which would be a more practical source of electricity for a car: a wet cell or a dry cell? Why?
Finger [1]

Answer:

wet

Explanation:

4 0
2 years ago
N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t
son4ous [18]

Answer : The mole fraction of nitrogen will be 0.4615.


Explanation : When nitrogen (N_{2})and hydrogen (H_{2})are mixed, the mole ratio becomes 1 : 1.5,


Now we know that (H_{2}) is acting as a limiting agent.


So at the time of when 0.4 moles of (NH_{3}) is been formed it requires 0.4 moles of (N_{2}) and 3.4 moles of (H_{2})


So, we find the the remaining (N_{2}) will be 0.6 and

(H_{2}) will be 0.3 mole present in mixture.


So, the mole fraction of (N_{2}) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615

8 0
2 years ago
Read 2 more answers
If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?
Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

 

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

 

Answer:

<span>0.2 moles O</span>

8 0
2 years ago
Read 2 more answers
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