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3 years ago

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