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3 years ago

Is NaCl CH3OH LiOH or H2SO4 a bronsted acid

Chemistry

1 answer:

vesna_86 [32]3 years ago

3 0

Answer:

D. H₂SO₄

Explanation:

Bronsted acids are those that donate H+ ions. In this question, H₂SO₄ is a Bronsted acid.

Note: H₂SO₄ is one of seven strong acids that you should try to memorize.

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Which tool or tools would you use to measure and compare the mass of a cup of fresh water and a cup of salt water

Artemon [7]

You could use a scale to measure the mass as well as a cup to hold the water. If you were comparing the two, you should also probably use a graduated cylinder to get the same amount of each type of water.

Hope this helped ^_^

4 0

2 years ago

If water is added to a 0.70 molar solution of CuSO4 what will change

lana [24]

Answer:

B. Molarity will decrease

Explanation:

Molarity is one of the measures of the molar concentration of a solution. It is calculated by dividing the number of moles of the solute by the volume of the solvent. This means that the higher the amount of solute in relation to the volume of solvent, the higher the molarity of that solution.

In essence, adding water to a solution dilutes it i.e it increases the solvent's volume in relation to the solute, causing the molarity to decrease. In a nutshell, diluting a solution (by adding water or more solvent) causes the molarity of such solution to decrease. For example, if water is added to a 0.70 molar solution of CuSO4, the molarity of the solution will DECREASE.

7 0

2 years ago

The majority of the elements essential to life are found in what part of the periodic table?

IRINA_888 [86]

The rows in the top third - This group consists of elements like Sodium, Magnesium, Potassium and Calcium on the right and Chlorine, Carbon, Nitrogen and Oxygen on the left.
Sodium and Chlorine are components of salt, a very important compound of our blood, essential for transferring electrical signals from the brain to the rest of the body and vice versa. Calcium is the building block of our bones, while Magnesium and potassium ensure proper functioning of our organs.

3 0

2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

Which element is a halogen? argon bromine calcium lithium

almond37 [142]

It is (CI) bromine
because, Halogen element, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts).

6 0

2 years ago

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