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puteri [66]
3 years ago
10

Science notation for 5,098,000

Chemistry
2 answers:
3241004551 [841]3 years ago
8 0
You simply move the decimal six places to the left
Aleksandr [31]3 years ago
4 0
5.098x10^6
You simply move the decimal six places to the left to make the decimal land after 5.
Hope this helps!
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What is the mass of oxygen gas in a 16.8 l container at 23.0◦c and 2.50 atm? answer in units of g?
FromTheMoon [43]
Assuming ideality:

PV=nRT

Pv= (mass/molarmass)RT

Solve for the mass :-)
4 0
3 years ago
in a chemical reaction known as decomposition, carbonic acid breaks down into water, and what other compound?
BartSMP [9]

Answer: it’s carbon dioxide

Explanation:

This is it because decomposition takes place when carbonic acid breaks down So that’s why I put carbon dioxide

8 0
3 years ago
Read 2 more answers
How many grams of CO are needed to react with an excess of Fe2O3 to produce (210.3 g Fe) 209.7 g Fe? Show your work.
Fed [463]

Explanation:

3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)

1)Mass of CO when 210.3 g of Fe produced.

Number of moles of Fe in 210.3 g=

\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}

=\frac{210.3}{55.84 g/mol}=3.76 mol

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : \frac{3}{2}\times 3.76 moles of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

\text{Number of moles}\times \text{molar mass of CO}=5.46\times 28 g/mol=157.92 g

2)Mass of CO when 209.7 g of Fe produced.

Number of moles of Fe in 209.7 g=

\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}

=\frac{209.7}{55.84 g/mol}=3.75 mol

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : \frac{3}{2}\times 3.75 moles of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

\text{Number of moles}\times \text{molar mass of CO}=5.625\times 28 g/mol=157.5 g

4 0
3 years ago
Read 2 more answers
You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. Aft
olya-2409 [2.1K]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

Mass=Density\times Volume

Volume of solution = Volume of HCl + Volume of Ba(OH)_2

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

Mass=1g/mL\times 93.1mL=93.1g

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat released = ?

m = mass = 93.1 g

C_p = specific heat capacity of water = 4.184J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 29.83^oC

Now put all the given value in the above formula, we get:

Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC

Q=1881.43J=1.88kJ        (1 kJ = 1000 J)

Now we have to calculate the moles of Ba(OH)_2 and HCl.

\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}

\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol

and,

\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}

\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

From the balanced reaction we conclude that

As, 1 mole of Ba(OH)_2 react with 2 mole of HCl

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and Ba(OH)_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2O

From the reaction, we conclude that

As, 1 mole of Ba(OH)_2 react to give 2 mole of H_2O

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of H_2O

Now we have to calculate the change in enthalpy of the reaction.

\Delta H_{rxn}=-\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=-\frac{1.88kJ}{0.0194mole}=-96.9kJ/mole

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

3 0
3 years ago
Hello this is my 3 attempt, please someone answer this i'm begging you.
denis23 [38]

Answer:

Explanation:

No worries, I got you :)

So for the first question, you need to use PV=nRT to find the n, or in other words, the number of moles. Then, you can find the molar mass since you know the grams and the moles

110 kPa / 101.3 = 1.085 atm (I converted it to atm so I can use the .08206 L atm/ k mol  for the rate)

550 ml / 1000 = .550 L (I converted mL to L in order to use the .08206 L atm/ k mol for the Rate)

28.5 c + 273 = 301.5 K (I converted C to K in order to use the .08206 L atm/ k mol for the Rate)

PV=nRT

(1.085) (.550 L) = n (0.08206) (301.5)

Divide the (0.08206) (301.5) to get n alone:

(1.085) (.550 L) / (0.08206) (301.5) = n

When I divided, I got n= .02412 moles, and since we have 1.88g , we divide the 1.88 by .02412 to get the molar mass (grams/mole)

77.94 g/mole is the molar mass

we know that there are 3 H's in the compound, so we do 3(1.008) and subtract 77.94 by what you get.

3 x 1.008 = 3.024 -----> 77.94-3.024 = 74.9

Now we look at the periodic table and try to find an element that has a molar mass of 74.9

Arsenic (As) has a molar mass of 74.922, which is close enough. Plus, Arsenic has a charge of 3, so it fits with the 3 hydrogens.

5 0
3 years ago
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