What is the empirical formula for a compound if a sample contains 1.0 g of S and 1.5 g of O?

What kind of reaction does Iron (III) and copper (II) sulfate solution produce?

ivanzaharov [21]

Answer:

The answer to your question is Single replacement

Explanation:

Types of reaction in Chemistry

Synthesis two or more reactants combine to produce one single product

Decomposition one reactant splits producing two or more products

Single replacement One element replaces a similar one in a compound.

Double replacement Two compounds interchange their cations and anions.

Combustion one compound reacts with oxygen to produce carbon dioxide and water.

Reaction given

Fe⁺² + CuSO₄ ⇒ Cu + FeSO₄

This is a single replacement reaction

4 0

2 years ago

Read 2 more answers

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B

natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

$A = A_{0}e^{-kt}$

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

$ln[A] = ln[A]_{0}-kt$

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

7 0

3 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0

3 years ago

Rate how successful you were at using Active Reading skills by responding to the following statement: As i read, i paid attentio

Aleksandr-060686 [28]

It is possible to rate your active reading skills either by using a number is a specific rate or by using a word that qualifies your performance.

<h3>What are active reading skills?</h3>

Active reading skills are a set of skills that help you read actively; this means skills to effectively read a text and understand it.

<h3>How to rate your reading skills?</h3>

In this case, you are asked to rate your ability to pay attention to your understanding, thoughts, and questions while reading. To do this you can:

Rate using numbers:

Select a specific scale rating such as 1 to 5 or 1 10.
Analyze honestly if you used this ability while reading the text.
Give yourself a number. For example, if you apply this skill only sometimes while reading you can rate yourself 5 out of 10.

Rate using words:

Select the types of words that you will use, the most common ones are frequency words such as always, sometimes, and never.
Analyze honestly your performance.
Find the word that best fits it.

Learn more about rate in: brainly.com/question/2278897

8 0

1 year ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

2 years ago