Please fill in the blanks:

Convert 35 g of iron to moles of iron.

Ad libitum [116K]

Answer:

ummmmm 35???????

Explanation:

5 0

2 years ago

Which of the following is a reactant in photosynthesis?

svet-max [94.6K]

Answer:

This would be Water

Explanation:

The reactants for photosynthesis is light or energy from the sun, carbon dioxide which is what plants breath, and chlorophyll for channeling the energy of the sun into chemical energy.

the rest of the answers is the products of photosynthesis not the reactants.

If you have any questions feel free to ask in the comments - Mark

ALSO Happy Valentines Day ❤

7 0

3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

3241004551 [841]

Answer: " 13.5 g Al " ;

→ that is: "13.5 grams of aluminum."

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Explanation:

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Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

The reactants: "aluminum (Al) " ; and "chlorine (Cl) " ; and:

The product: "aluminum choloride (AlCl₃) " .

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The "balanced chemical equation" is:

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2 Al + 3 Cl₂ → 2 AlCl₃ ;

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Note: The molecular weight of "aluminum (Al)" is: " 26.98 g /mol " .

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So: We call solve using a technique known as: "dimensional analysis" :

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0.500 mol AlCl₃ * $(\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?$

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Note: The units of "mol AlCl₃" cancel out to "1' ; and:

The units of "mol Al" cancel out to "1" ; and we are left with:

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" $\frac{(0.500 * 2 * 26.98)}{2}$ g Al ["grams of aluminum"] ;

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Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

→ (0.500 * 26.98) g Al ;

= 13.49 g Al ;

→ Round to 3 (Three) significant figures;

→ Since: "0.500" has 3 (Three) significant figures:

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= 13.5 g Al ; that is: "13.5 grams of aluminum."

____________________________

Hope this is helpful!

Best wishes to you in your academic pursuits—and within the "Brainly" community!

____________________________

3 0

3 years ago

A chemistry instructor provides each student with 8 test tubes at the beginning of the year. If there are 28 students per class,

klemol [59]

Three classes: 28×3=84 students
8 test tubes per student: 84× 8= 672
hope this helps!

7 0

3 years ago