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2 years ago

Match the following vocabulary words with the correct definitions.

Chemistry

1 answer:

son4ous [18]2 years ago

4 0

Answer:

1 mass

2 matter

3 weight

4 molecule

5 compound

6 atom

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Si2Br6 compound name

Anna71 [15]

Answer:

disilicon hexabromide

Explanation:

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2 years ago

Convert the scientific notation to a decimal number. Type the correct answer in the box. 4 × 10-5 cm = cm

MakcuM [25]

Answer:

$4 * 10^{-5}cm = 0.00004cm$

Explanation:

Given

$4 * 10^{-5}cm = [\ ]cm$

Required

Solve

Apply law of indices

$4 * \frac{1}{10^5}cm = [\ ]cm$

Evaluate 10^5

$4 * \frac{1}{100000}cm = [\ ]cm$

$4 * 0.00001cm = [\ ]cm$

$0.00004cm = [\ ]cm$

Hence:

$4 * 10^{-5}cm = 0.00004cm$

3 0

3 years ago

Wat are representative elements

AnnyKZ [126]

In chemistry and atomic physics, the main group is the group of elements whose lightest members are represented by helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, and fluorine as arranged in the periodic table of the elements

7 0

2 years ago

Read 2 more answers

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago

What is the parent isotope when Cl-33 is formed during a beta decay?

zzz [600]

Answer:

33/16 S

Explanation:

In beta decay, the atomic number of the daughter nucleus increases by one unit while the mass of the daughter nucleus remains the same as that of the parent nucleus.

Hence, if we know that a beta decay has occurred, then the parent nucleus must have the same mass as its daughter nucleus but have an atomic number that is less than that of the daughter nucleus by only one unit, hence the answer above.

7 0

3 years ago