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2 years ago

8. What must occur for a change to be a chemical reaction?

Chemistry

1 answer:

Maslowich2 years ago

7 0

Answer: A. there must be a change in chemical properties

Explanation:

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Use the References to access important values if needed for this question There are 12 eggs in a dozen. If a farmer's chickens p

bekas [8.4K]

Answer:

$3.4\times 10^3 eggs$ are produces are in a month.

Explanation:

Quantity of eggs produced by the chicken in a month = 284 dozens

1 dozen = 12 eggs

Number of eggs in a month:

$284 dozens = 284\times 12 eggs =3,408 eggs\approx 3.4\times 10^3 eegs$

$3.4\times 10^3 eggs$ are produces are in a month.

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2 years ago

Pls this need this I am doing a test and I need to get this

Alexxx [7]

Answer:

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Explanation:

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2 years ago

A sample of argon gas collected at a pressure of 0.918 atm and a temperature of 279 K is found to occupy a volume of 719 millili

babymother [125]

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3 years ago

What mass of oxygen is needed for the complete combustion of 4.60×10−3g of methane?

sp2606 [1]

First step in answering the question is to establish a balanced chemical reaction equation. More specifically, a combustion chemical equation.

CH4 + 2O2 ---> CO2 + 2H20

Then using dimension analysis:

$4.60*10^{-3} g CH4 ( \frac{moleCH4}{16 g CH4}) * ( \frac{2mole O2}{mole CH4}) * ( \frac{32 g O2}{mole O2} ) = 0.0184 g O_{2}$

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3 years ago

What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium

alex41 [277]

Answer:

Approximately $3.06 \times 10^{2}\; \rm g$ (approximately $306\; \rm g$ .)

Explanation:

Calculate the quantity $n$ of lithium phosphate in $V = 2.5\; \rm L$ of this $c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}$ lithium phosphate solution.

$\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}$ .

Empirical formula of lithium phosphate: $\rm Li_3PO_4$ .

Look up the relative atomic mass of $\rm Li$ , $\rm P$ ,and $\rm O$ on a modern periodic table:

$\rm Li$ : $6.94$ .
$\rm P$ : $30.974$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm Li_3PO_4$ :

$M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}$ .

Calculate the mass of that $n = 2.65\; \rm mol$ of $\rm Li_3PO_4$ formula units:

$\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}$ .

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2 years ago