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Doss [256]
2 years ago
15

One peanut M&M weighs approximately 2.33 g.

Chemistry
1 answer:
Sphinxa [80]2 years ago
3 0

Answer:

There are 23076 peanut M&M's in 53.768 kg of M&M's.

Explanation:

First we <u>convert 53.768 kg into g</u>:

  • 53.768 kg * 1000 = 53768 g

Then we <u>divide the total mass of M&M's by the mass of one peanut M&M,</u> in order to calculate the answer:

  • 53768 g / 2.33 g = 23076

So there are 23076 peanut M&M's in 53.768 kg of M&M's.

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Convert 27.7 kilometers to centimeters.
Stolb23 [73]

Answer:

The correct answer is - 2770000 cm.

Explanation:

1 kilometer = 1000 meter

1 meter = 100 centimeter

1 kilometer = 100*1000 cm

1 km = 100000 cm.

then,

27.7 kilometers = 2.77 × 10^6 centimeters

So, 27.7 kilometers = 27.7 × 100000

= 2.77 × 106 or 2770000 centimeters.

7 0
2 years ago
Which phase change is exothermic?
DerKrebs [107]
Freezing, condensation, Deposition.
8 0
3 years ago
What is true about the position of any metal in the periodic table
Serjik [45]

Answer:

The metals are on the bottom left in the periodic table, and the nonmetals are at the top right. The semimetals lie along a diagonal line separating the metals and nonmetals. The elements are arranged in a periodic table, which is probably the single most important learning aid in chemistry.

Explanation:

im not sure if there was supposed to options to that question or not

7 0
3 years ago
1pt The heaviest concentration of asteroids lies in a belt between
Olegator [25]
The answer is A, Mars and Jupiter.
7 0
3 years ago
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
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