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3 years ago

Which substance can not be decomposed by a chemical change?

Chemistry

2 answers:

Dvinal [7]3 years ago

7 0

Copper. It's copper because it is an element, not a compound substance, so its already in its basic chemical form ^u^

max2010maxim [7]3 years ago

5 0

B. Copper since the rest of the choices are compounds. Copper is an element.

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Bas +<br> PtF2 →<br> BaF2 +<br> Pts<br> Need to balance it

Basile [38]

it is already balanced

REACTANTS

Barium sulfide (BaS) + platinum (Ii) fluoride

PRODUCT

Barium fluoride (BaF2) + Cooperite (PtS)

Hope this answer helps you dear! take care

5 0

2 years ago

Write the difference between the reaction of Calcium and Aluminium with water.

guapka [62]

Answer:

Both gives hydroxides, but

Calcium reacts with water to give calcium hydroxide Ca(OH)2. this product is slightly soluble in water.

Aluminium is less reactive than Ca, so it reacts with only steam, to give aluminium hydroxide Al(OH)3. This product is insoluble in water.

5 0

3 years ago

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

3 years ago

A sample of gas has a volume of 1.40 L at a pressure of 225 mmHg. What is the new volume at a pressure of 515 mmHg?

guajiro [1.7K]

Answer:0.61L

Explanation:

P1V1/n1T1=P2V2/n2T2

n1=n2 T1=T2

Therefore;

P1V1=P2V2

225×1.4=515×V2

V2=0.61L

N.B: Increase in pressure leads to decreases in volume (Boyle's law)

5 0

3 years ago

Photo attached, please help me if you can :(

mylen [45]

Answer:

I'm pretty sure the ionic structure is C, sorry this is only one answer tho : , (

Explanation:

7 0

3 years ago