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ANTONII [103]
2 years ago
12

What formula is used for solving problems involving Boyle’s law?

Chemistry
1 answer:
GaryK [48]2 years ago
5 0

Answer:

P1V1=P2V2

Explanation:

because volume is inversely proportional to pressure provided that temperature remains constant

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Which of these is an oxidation half-reaction?
Zepler [3.9K]

Answer:

E= mc sqared

Explanation:

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6 0
4 years ago
Draw structural formulas for all the alkene(s) formed by treatment of each haloalkane or halocycloalkane with sodium ethoxide in
poizon [28]

Answer:

Explanation:

Kindly note that I have attached the complete question as an attachment.

Here, we are told that elimination occurs by an E2 mechanism. What this means is that the hydrogen and the halogen must be above and below for the reaction to proceed.

The possible products are as follows;

Please check attachment for complete equations and diagrams of compounds too.

5 0
3 years ago
What is Gram Atomic mass ?​
KIM [24]

Answer:

Gram atomic mass of an element can be defined as the mass of one mole of atoms of a particular element. It is numerically equivalent to the value of the element's atomic mass unit but has its unit in grams.

5 0
2 years ago
Read 2 more answers
Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma
Elena-2011 [213]

Answer:

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

Explanation:

CaCO_3(s)\rightarrow CaO(s) + CO_2(g)

Annual production of CaO = 8.6\times 10^{10} kg=8.6\times 10^{13} g

Moles of CaO :

\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with  1.53\times 10^{12} moles of CaO moles of carbon-dioxide moles produced will be:

\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles of carbon-dioxide

Mass of 1.53\times 10^{12} moles of carbon-dioxide:

1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

6 0
3 years ago
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde
nadya68 [22]

Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=2.74^0C = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}

M=200g/mol

Thus the molar mass of the unknown compound is 200 g/mol

4 0
4 years ago
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