All categories

3 years ago

15 POINTS - AUXILARY LINE

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

8 0

An auxiliary line (or helping line) is an extra line needed to complete a proof in plane geometry. Other common auxiliary constructs in elementary plane synthetic geometry are the helping circles.

You might be interested in

ASAP PLEASE

marishachu [46]

Answer:

Explanation:

the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.

8 0

3 years ago

Read 2 more answers

Nitrogen dioxide gas is produced by the reaction of nitrogen monoxide gas and oxygen gas .

tensa zangetsu [6.8K]

2NO + O2 ----> 2NO2

there ya go

8 0

3 years ago

A 16 gram sample of O2(g) fills a container at STP.<br> What volume is the container?

sergij07 [2.7K]

Answer:

V = 11.2 L

Explanation:

Hello there!

In this case, according to the ideal gas equation:

$PV=nRT$

It is possible to compute the volume as shown below:

$V=\frac{nRT}{P}$

Whereas the moles are computed are computed given the mass and molar mass of oxygen:

$n=16g*\frac{1mol}{32g} =0.5mol$

Now, since the STP stands for a temperature of 273.15 K and a pressure of 1 atm, the resulting volume is:

$V=\frac{0.5mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=11.2L$

Best regards!

8 0

3 years ago

How many atoms of gold (Au) are equal to 212g of gold?

Lostsunrise [7]

Given :

Mass of gold ( Au ) is 212 gram.

To Find :

Atoms of gold present in 212 gram of gold.

Solution :

Molecular mass of gold is 197 gram.

So, number of moles of gold in 212 gram is :

$n = \dfrac{212}{197} \ moles\\\\n = 1.076 \ moles$

Now, we know 1 mole of any element contains $6.022 \times 10^{23}$ atoms.

So, number of atoms present in 1.076 moles are :

$N = 1.076 \times 6.022\times 10^{23}\\\\N = 6.48 \times 10^{23}$

Hence, this is the required solution.

8 0

3 years ago

1 mole of no2(g) has a greater entropy than 1 mole of n2o4(g). true or false

bija089 [108]

In order to answer this, you need to find the empirical data for the standard entropies. Please refer to this link: http://www.mrbigler.com/misc/energy-of-formation.PDF

For NO₂ gas, the entropy is 240 J/mol-K. For N₂O₄ gas, the entropy is 304.2 J/mol-K. Therefore, <em>the statement is false.</em>

7 0

3 years ago