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4 years ago

What is 30-9x2*2-21-4+4

1 answer:

5 0

Answer:
9 (1-2x²)

Explanation:
The given expression is:
30 - 9x²*2 - 21 - 4 + 4
The first step is to compute the multiplication. This will give:
30 - 18x² - 21 - 4 + 4
Then, we will add like terms as follows:
(30-21-4+4) - 18x²
= 9 - 18x²
Finally, we can take the 9 as a common factor from both terms, this will give:
9 (1-2x²)

Hope this helps :)

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3 years ago

Sodium and chloride ions readily combine because they have opposite charges.

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3 years ago

What is the percent composition of N H S O in (NH4)2SO4

Lena [83]

Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

- 2 moles of N.

- 8 moles of H.

- 1 mole of S.

- 4 moles of O.

2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

- 2 moles of N (14g/mol) = 28g

- 8 moles of H (1g/mol) = 8g

- 1 mole of S (32g/mol) = 32g

- 4 moles of O (16g/mol) = 64g

3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

$\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}$ $\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}$ $\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}$ $\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\ \\ \end{gathered}$

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

8 0

1 year ago

HELP PLEASEEEE

kifflom [539]

Explanation:

So first find out the moles. The percentage is out of 100 so just convert the percentage into mass by just changing the sign.

Ga = 55g/69.7 = 0.789 moles

F=45g/19 = 2.37 moles

Then divide by smallest value to find out the formula

Ga=0.789÷0.789= 1

F=2.37÷0.789 = 3

GaF3

hope this helps:)

6 0

2 years ago

3. Heating the copper product at too high a temperature in an oxygen atmosphere results in the formation of copper( )oxde, . Wri

olya-2409 [2.1K]

First, write the equation:
Cu + O2 -> CuO
Now, balance:
2 Cu + O2 -> 2 CuO

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3 years ago