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timama [110]
3 years ago
5

Which of the following collection is the more positive view of life and society in Blake's time?

Chemistry
2 answers:
o-na [289]3 years ago
8 0

Answer : The correct answer is a) Songs of Innocence .

Blake's songs of Innocence and experience juxtaposes the innocence of childhood and harsh realities and corruption of adult hood .

Song of Innocence says about the hopes and fears of childhood and the thoughts of child as adults . It draws attention towards positive aspect of human before entering into adulthood world of corruption.

While songs of experience states about corrupt world and destroyed innocence of adulthood . It also shows repressive effect like jealousy , shame , secrecy etc .

Hence we can song of innocence shows more positive aspect of life .

Reptile [31]3 years ago
4 0
The anser is c hope this helps 
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5 0
3 years ago
Choose the molecular compound among the substances listed below.
agasfer [191]

Answer: Among the listed substances XeCl_{4} is the molecular compound.

Explanation:

A chemical compound formed by the chemical combination of two or more non-metals is called a molecular compound or covalent compound.

For example, Xe and Cl are non-metals. The compound formed by them is XeCl_{4} which is a molecular compound.

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4 0
3 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

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