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Charra [1.4K]
3 years ago
10

Which sentence from “Broken Chain” contains the most descriptive language? “He asked his mother if he could have braces, like Fr

ankie Molina, her godson, but he asked at the wrong time.” “He went to the back yard, where he unlocked his bike, sat on it with the kickstand down, and pressed on his teeth.” “Sunday morning, Ernie and Alfonso stayed away from each other, though over breakfast they fought over the last tortilla.” “At four he decided to get it over with and started walking to Sandra’s house, trudging slowly, as if he were waist-deep in water.”
Chemistry
2 answers:
Molodets [167]3 years ago
8 0

THE ANSWER IS D.  I GOT IT RIGHT ON THE CUMULATIVE EXAM REVIEW

Sergeeva-Olga [200]3 years ago
3 0

Answer:

" At four he decided to get it over with and started walking to Sandra's house, trudging slowly, as if he were waist-deep in water. "

Explanation:

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What do these substances have in common?
Akimi4 [234]

They are compounds Good Luck!

4 0
3 years ago
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Complete combustion of a 17.12mg sample of xylene In oxygen yielded 56.77mg
Veronika [31]

Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=

106.16×1000

17.12

=0.00016moles

Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO

2

=

44.01×1000

56.77

=0.0013

Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H

2

O==

18.02×1000

14.53

=0.0008

Moles ratios

\frac{0.0013}{0.0008}=1.625

0.0008

0.0013

=1.625

\frac{0.0008}{0.0008}=1

0.0008

0.0008

=1

Hence molecular fomula

The empirical formula is C 4H 5.

The molecular formula C8H10

8 0
2 years ago
Li+1 had gained ______ one electron
zysi [14]
Maybe molecules one electron
3 0
3 years ago
PLEASE HELP QUICK I WILL GIVE BRAINLIEST
Dvinal [7]

Answer:

3.00 moles Mg

Explanation:

4 0
3 years ago
First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
Klio2033 [76]

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0
2 years ago
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