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3 years ago

1. Mixing is a physical change. True or false

Chemistry

2 answers:

Arada [10]3 years ago

6 0

Answer:

false it is a chemical change

Hitman42 [59]3 years ago

5 0

Answer:

true

Explanation:

if u mix it can explode which is a chemical reaction

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the speed limit is posted as 35 km/hr . your speedometer reads that you are going 40 miles/hr. are you speeding?

Morgarella [4.7K]

No, because 40 miles is the same as nearly 25 km/h.

4 0

3 years ago

The molar mass of I2 is 253.80 g/mol, and the molar mass of NI3 is 394.71 g/mol. How many moles of I2 will form 3.58 g of NI3?

Nat2105 [25]

To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.

6 0

3 years ago

What is the mass of 2.071x10^14 molecules of SF2

kolezko [41]

Answer:

Name of molecule Sulfur Difluoride ( SF2)

No of Valence Electrons in the molecule 20

Hybridization of SF2 sp3 hybridization

Bond Angles 98 degrees

Molecular Geometry of SF2 Bent

Explanation:

Sulfur Difluoride is an inorganic molecule made up of one Sulphur atom and two Fluorine atoms. It has a chemical formula of SF2 and can be generated by the reaction of Sulphur Dioxide and Potassium Fluoride or Mercury Fluoride. In this blog post, we will look at the Lewis dot structure of SF2, its molecular geometry and shape.

7 0

3 years ago

Which metal cations form a white precipitate with both H2SO4(aq) and Na2S2O4(aq)? 1. Ba2+ 2. Sr2+ 3. Fe2+ 4. Cu2+

8090 [49]

Answer:

1. Ba2+ 2. Sr2+

Explanation:

When a solution contains the Barium ,Ba²⁺ ion or Strontium, Sr²⁺ ion, they reacts with either H₂SO₄(aq) or Na₂SO₄(aq) to produce a white precipitate of BaSO₄(s) and SrSO₄(s) respectively

The chemical reactions are given below

Ba²⁺ + H₂SO₄(aq) ⇒ BaSO₄(s) + 2H⁺ (aq)

Ba²⁺ + Na₂SO₄(aq) ⇒ BaSO₄(s) + 2Na⁺ (aq)

Sr²⁺ + H₂SO₄(aq) ⇒ SrSO₄(s) + 2H⁺ (aq)

Sr²⁺ + Na₂SO₄(aq) ⇒ SrSO₄(s) + 2Na⁺ (aq)

6 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