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3 years ago

Could someone please help on just question 2?

Chemistry

1 answer:

sp2606 [1]3 years ago

4 0

Hey,

A compound is a substance that is made of two or more type of atoms chemically bonded together.
So this means that they are joined together by a chemical reaction.
And of those options PF3 H2SO4 H2O And CO are compounds as you can see they are made of two different type of atoms

Hope this helps !!

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How many moles of MgCl2 are there in 340 g of the compound?

iragen [17]

To calculate the amount of the compound in the units of grams, we need to first obtain for the molar mass of the compound. We calculate as follows:

MgCl2 = 24.31 + (2 x 35.45) = 95.21 g/mol

340 g MgCl2 ( 1 mol / 95.21 g ) = 3.57 mol MgCl2

Hope this answers the question. Have a nice day.

7 0

3 years ago

Read 2 more answers

A 2.9 kg model rocket accelerates at 15.3 m/s2 with a force of 44 N. Before launch, the model rocket was not moving. After the s

masha68 [24]

Answer:

Newton's first and third law of Motion

Explanation:

The laws applying in the example Newton's first and third laws of Motion.

The first law states that any object at rest (ie. not moving) will stay at rest until it is forced to move by an external force. In this case, said force were the propulsion gases ignited.

As the hot gases were pushed out from the engine nozzle, there was another force equal in magnitud but opposite in direction (as the gases went down, that force went upwards), said force is directly responsible for the rocket taking off. That is an example of the third law.

7 0

3 years ago

The reaction H2O+CH3Br=CH3OH+HBr. H2O acts as what

uysha [10]

Answer:

H2O acts as an oxidizing agent

3 0

2 years ago

How many grams are in 1.76 x 10^23 atoms of iodine

Mariana [72]

Answer:

$\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}$

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

$6.022*10^{23}$

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

$\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Multiply the given number of atoms by Avogadro's number.

$1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Flip the fraction so the atoms of iodine will cancel out.

$1.76*10^{23} \ atoms \ I*\frac{ 1 \ mol \ I}{6.022*10^{23} \ atoms \ I}$

$1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }$

Multiply so the problem condenses into 1 fraction.

$\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }$

$0.2922617071 \ mol \ I$

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

$\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply this fraction by the number of moles found above.

$0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply. The moles of iodine will cancel.

$0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }$

The 1 as a denominator is insignificant.

$0.2922617071 *{ 126.9045 \ g\ I }$

$37.08932581 \ g \ I$

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

$37.08932581 \ g \ I$

The 8 in the hundredth place tells us to round the 0 up to a 1.

$\approx 37.1\ g \ I$

There is about 37.1 grams of iodine in 1.76*10^23 atoms.

5 0

3 years ago

What do the observations for changes to volume and pressure show you about the

Irina-Kira [14]

Answer: The given demonstration does not support Boyle's Law.

Explanation:

According to Boyle's law, at constant temperature the pressure of an ideal gas is inversely proportional to the volume of gas.

$P \propto \frac{1}{V}$

Therefore, an increase in volume of gas will lead to a decrease in pressure and vice-versa.

Now, according to the given data of volumes there is occurring an increase on moving from final volume 1 to final volume 3. This means that pressure is decreasing.

As a result, if a balloon was squeezed or compressed then it would get smaller. But in the given demonstration pressure is not given.

Thus, we can conclude that the given demonstration does not support Boyle's Law.

3 0

3 years ago