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2 years ago

I WILL GIVE U BRAINLY !!!!The diagram shown is an example of the element

Chemistry

2 answers:

Eduardwww [97]2 years ago

4 0

Answer:

It is boron

Explanation:

Because the image shows only 5 protons, the element can be identified as the element with the atomic number 5. This element is boron.

motikmotik2 years ago

3 0

Your answer is a, boron. Boron has 5 protons, 6 neutrons, and 5 electrons. Hope this helps!

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Which of the following are the starting substances in photosynthesis?

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A: water and carbon dioxide.

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3 years ago

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Which condition must be met in order for an equation to be balanced?

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Answer:

The elements in the reactants are the same as the elements in the products.

Explanation:

You can't have more or less elements or atoms, but you also can't just have the same number of atoms because then it could be different elements, which can't happen.

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What does it mean for heat to be transferred by radiation?

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Answer:

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Explanation:

It means that the heat comes directly at you without relying on any material to conduct it.

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3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

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3 years ago

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Determine the theoretical maximum moles of ethyl acetate, , that could be produced in this experiment. The reactant, acetic acid

kherson [118]

Answer:

0.1832 moles of ethyl acetate ( $C_{4}H_{8}O_{2}$ )

Explanation:

1. Find the balanced chemical equation:

In the production of ethyl acetate, the acetic acid $CH_{3}COOH$ reacts with ethanol to produce ethyl acetate $C_{4}H_{8}O_{2}$ and water, that is:

$CH_{3}COOH+C_{2}H_{5}OH=C_{4}H_{8}O_{2}+H_{2}O$

2. Find the theoretical maximum moles of ethyl acetate $C_{4}H_{8}O_{2}$ :

As the problem says that the acetic acid $CH_{3}COOH$ is the limiting reagent, use stoichiometry to find the moles of ethyl acetate produced:

$11gCH_{3}COOH*\frac{1molCH_{3}COOH}{60.05gCH_{3}COOH}*\frac{1molC_{4}H_{8}O_{2}}{1molCH_{3}COOH}=0.1832molesC_{4}H_{8}O_{2}$

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3 years ago