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k0ka [10]
3 years ago
13

Which of the following describes an atom?

Chemistry
1 answer:
vredina [299]3 years ago
7 0

Answer:

Correct answer is option (3) .

Explanation:

Hope it helpful....

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How many grams of oxygen gas must react to give 2.10g of ZnO?
Advocard [28]
The chemical reaction is written as:

2Zn + O2 = 2ZnO

We are given the amount of the product to be produced from the reaction. We use this value and the relation of the substances in the reaction to calculate what is asked. We do as follows:

2.10 g ZnO ( 1 mol / 81.408 g ) ( 1  mol O2 / 2 mol ZnO ) ( 32 g / 1 mol ) = 0.414 g O2 is needed
8 0
4 years ago
68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi
Vera_Pavlovna [14]

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>

Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:

  • 0.528 mol Mg(NO₃)₂ * \frac{1molMg(OH)_2}{1molMg(NO_3)_2} = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

  • 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g
7 0
3 years ago
Different measurements are expressed in different units. Choose the correct Sl units for the lo
swat32

Answer:

The SI units for measuring the velocity of the car: meters per second (m/s)

The Sl units for measuring the acceleration of the car: meters per second squared  (m/s^2)

The SI units for measuring force:  Newton (N)

The SI units for measuring mass: kilograms (kg)

Explanation:

Système international (SI) or International System of Units consist of a list of unit measurement that mostly used by scientist. The scientist from a different country might use different unit that makes them have to convert the result of each other. The usage of the same unit measurement will help the scientist to read that publication easier, make it easier to share and discuss any topic. The unit used is metric since the decimal system also makes the conversion of a unit easier.  

8 0
3 years ago
John was considered to be a pillar in the early church?
o-na [289]
Yes he was a pillar in the early church
4 0
4 years ago
Read 2 more answers
Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to
snow_lady [41]

Answer:

\displaystyle \text{p} K_a \approx 3.856

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:


\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right]  & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

5 0
2 years ago
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