A compound has an empirical formula of Na2CO3 and a molar mass of 318g/mol. Find its

CuCl2 + NaNO3 -------> Cu(NO3)2 + NaCl

pav-90 [236]

Answer:

what

Explanation:

3 0

2 years ago

A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

6 0

2 years ago

copper forms two oxides. On heating 1 g of each in hydrogen 0.888 g and 0.798 g of the metal was obtained. Show that the results

Mashutka [201]

Answer:

For the first oxide, 1 g gives 0.888 g of copper.

Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.

For the second oxide, 1 g gives 0.798 g of copper.

Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.

So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.

Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.

6 0

2 years ago

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Draw the structures of two stereoisomeric alkenes that would give 3-hexanol as the only major product of hydroboration followed

soldi70 [24.7K]

Answer:

cis-3-hexene and trans-3-hexene

Explanation:

Hydroboration oxidation is a method of preparation of alcohol from alkene.

Hydroboration follows anti-Markovnikoff rule in which alcohol group attached to less substituted carbon. Stereochemistry of the product is always syn that is H and OH attached to the same side of the double bond.

cis-3-hexene and trans-3-hexene undergoes hydroboration to form 3-hexanol.

7 0

2 years ago

How do I do this?<br> 100 points <br> Please help

harina [27]

Answer:

Explanation:

OK this is simple

i'll try to do the unanswered question

CaCl2

In the table

You look for calcium,Ca you fill 1

THen for Chloride=2

i hope you've understood

5 0

3 years ago

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