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3 years ago

8

In the reaction _S +3O2 → 2SO3, what coefficient should be placed in front of the S to balance the reaction? i need the answer.

sorry this isn't a chemistry question i need help with it
1

2

3

5

2 answers:

sukhopar [10]3 years ago

8 0

Answer:

you momma

Step-by-step explanation:

Bas_tet [7]3 years ago

4 0

5 siempre estoy corectooooooooooo

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A radioactive isotope has a half life of 4.51 days. If 12.63 grams are now present, how much will be left after 15

Marizza181 [45]

Answer:

Step-by-step explanation:

$f=ir^t\\ \\ \frac{1}{2}=r^{4.51}\\ \\ ln(0.5)=4.51lnr\\ \\ r=e^{\frac{ln0.5}{4.51}}\\ \\ r\approx 0.8575\\ \\ f(15)=12.63(0.8575^{15}\\ \\ f(15)\approx 1.26g$

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3 years ago

Sam bought a computer and printer for ₹ 1,15,499. The cost of the computer was ₹ 85,789. What was the cost of the printer?

Yuki888 [10]

Answer:

₹29,710

Step-by-step explanation:

We minus ₹ 85,789 from ₹ 1,15,499 to get ₹ 29,710

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4 years ago

The population of a town y years after 1970 is 32,070 + 1,524y.

Shalnov [3]

Answer:

A

Step-by-step explanation:

1524y: this is your slope.

Notice it is positive. Thismust mean that your population can only increase. Therefore, C and D are wrong. Now what is increasing? 1524 each year.

8 0

3 years ago

Read 2 more answers

Find the indicated nth partial sum of the arithmetic sequence.<br> −7, −3, 1, 5, . . ., n = 30

TiliK225 [7]

Step-by-step explanation:

Given the Arithmetic sequence

$-7, -3, 1, 5, . . .$

An arithmetic sequence has a constant difference d and is defined by

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n$

$-3-\left(-7\right)=4,\:\quad \:1-\left(-3\right)=4$

$\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$d=4$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=-7$

as

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=4\left(n-1\right)-7$ ∵ $d=4$

$\mathrm{Arithmetic\:sequence\:sum\:formula:}$

$n\left(a_1+\frac{d\left(n-1\right)}{2}\right)$

$\mathrm{Plug\:in\:the\:values:}$

$n=30,\:\space a_1=-7,\:\spaced=4$

$=30\left(-7+\frac{4\left(30-1\right)}{2}\right)$

$=30\left(58-7\right)$ ∵ $\frac{4\left(30-1\right)}{2}=58$

$=30\cdot \:51$

$=1530$ ∵ $\mathrm{Multiply\:the\:numbers:}\:30\cdot \:51=1530$

Therefore, the indicated nth partial sum of the arithmetic sequence is 1530.

ANOTHER METHOD

as

$a_n=4\left(n-1\right)-7$

n = 30

$\sum _{n=1}^{30}\:4\left(n-1\right)-7$

$=\sum _{n=1}^{30}4n-11$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \sum a_n+b_n=\sum a_n+\sum b_n$

$=\sum _{n=1}^{30}4n-\sum _{n=1}^{30}11$

as

$\sum _{n=1}^{30}4n=1860$

and

$\sum _{n=1}^{30}11=330$

so

$=1860-330$

$=1530$

7 0

4 years ago

Sylvia decided to purchase 4 points in order to lower her interest rate on her

swat32

I’m Pretty sure it’s d

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4 years ago

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