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4 years ago

How many significant figures should the result have 2.524 g (5.1 • 106 g)

Chemistry

1 answer:

kvasek [131]4 years ago

7 0

Answer:

Two

Step-by-step explanation:

In multiplication problems, your answer can have no more significant figures than the number with the fewest significant figures.

Start with the calculation inside parentheses:

5.1 × 10.6 = 54.060 000 00 (by my calculator)

There are three significant figures in 10.6, but only two in 5.1.

In this were a number you had to report, you would round to two significant figures and report the answer as 54.

54.060 000 00 → 54 (to two significant figures)

However, you must multiply this intermediate answer by 2.524.

Therefore, you multiply the number in the calculator by 2.524 and get

54.060 000 00 × 2.524 = 136.447 4400

NOW, you round to two significant figures and report the answer as 140.

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4 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

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Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

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First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

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

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

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3 years ago

Which of the following are not single-displacement reactions?

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Answer:

$\boxed{\text{B and C }}$

Explanation:

In a single-displacement reaction, one element exchanges partners with another element in a compound.

$\textbf{A. } \rm Fe + 2HCl \longrightarrow FeCl_2 + H_2$

This is a single-displacement reaction, because the element Fe exchanges partners with H in HCl.

$\textbf{B. } \rm KOH + HNO_3 \longrightarrow H_2O + KNO_3$

This is not a single-displacement reaction, because it is a reaction between two compounds.

This is a double displacement reaction in which the K⁺ and H⁺ cations change partners with the anions.

$\textbf{C. } \rm Na_2S + 2HCl \longrightarrow 2NaCl + H_2S$

This is not a single-displacement reaction. It is another double displacement reaction, in which the Na⁺ and H⁺ cations change partners with the anions.

$\textbf{D. } \rm Ca + 2HOH \longrightarrow Ca(OH)_2 + H_2$

This is a single-displacement reaction, because the element Ca exchanges partners with H in H₂O.

$\boxed{\textbf{B and C }}$ are not single-displacement reactions.

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