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Alex73 [517]
3 years ago
8

At room temperature, which of these are oxide minerals? Select all that apply.

Chemistry
2 answers:
DerKrebs [107]3 years ago
7 0
C. and d. are oxyminerals.
Klio2033 [76]3 years ago
3 0

Answer:  C.Al_{2}O_{3} and D.Fe_{3}O_{4}

Explanation: Oxide minerals are the minerals in which the oxide anion is bonded to the central metal atom. These are useful for industrial purposes.

Out of the given options, only two of them can be categorized as the oxide minerals.  

H2O and CO2 will not be considered as oxide minerals.

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A party balloon contains 5.50 x 10 ^22 atoms of helium gas. What is the mass, in grams of the
eduard

Answer:

\boxed {\boxed {\sf A. \ 0.365 \ g\ He}}

Explanation:

<u>1. Convert Atoms to Moles</u>

We must use Avogadro's Number: 6.022*10²³. This is the number of particles (atoms, molecules, ions, etc.) in 1 mole of a substance. In this case, the particles are atoms of helium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Multiply by the given number of helium atoms.

5.50 *10^{22} \ atoms \ He *\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Flip the fraction so the atoms of helium cancel.

5.50 *10^{22} \ atoms \ He *\frac {1 \ mol \ He}{6.022*10^{23} \ atoms \ He}

5.50 *10^{22} *\frac {1 \ mol \ He}{6.022*10^{23} }

\frac {5.50 *10^{22} \ mol \ He}{6.022*10^{23} }= 0.09133178346 \ mol \ He

<u>2. Convert Moles to Grams</u>

We must use the molar mass, which is found on the Periodic Table.

  • Helium (He): 4.00 g/mol

Use this as a ratio.

\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

Multiply by the number of moles we calculated. The moles will then cancel.

0.09133178346 \ mol \ He *\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

0.09133178346*\frac { 4.00 \ g \ He }{ 1 }

0.3653271338 \ g\ He

<u>3. Round </u>

The original measurement has 3 significant figures (5, 5, and 0). Our answer must have the same. For the number we calculated, it is thousandth place. The 3 in the ten thousandth place tells us to leave the 5.

0.365 \ g\ He

The mass is <u>0.365 grams of helium</u> so choice A is correct.

7 0
2 years ago
Describe how transpiration and respiration show interactions among Earth's system
stepladder [879]

Earth has 4 systems/ spheres which are known as Lithosphere(land), Biosphere(living things), Hydrosphere(water) and Atmosphere(air).

Transpiration is the process that interact with the Hydrosphere and Respiration is the process that interacts with the Atmosphere.

Transpiration is the process happening in plants which absorb and distribute water through their roots and release water in the form of vapors through their leaves. As this process involves Water, therefore this process interacts with the Hydrosphere.  

Respiration is the process in which plants convert carbon dioxide to oxygen as a part of photosynthesis which was then inhaled by the animals. As this process involves Air, therefore this process interacts with the Atmosphere.

7 0
3 years ago
Read 2 more answers
In the reaction Mg+Cl2=MgCl2 what should be the coefficient for Cl2
oksano4ka [1.4K]
It should just be one for it to be balanced.
7 0
3 years ago
Read 2 more answers
If 20.0g of CO2 and 4.4g of CO2
Ksivusya [100]

The given question is incorrect. The correct question is as follows.

If 20.0 g of O_{2} and 4.4 g of CO_{2}  are placed in a 5.00 L container at 21^{o}C, what is  the pressure of this mixture of gases?

Explanation:

As we know that number of moles equal to the mass of substance divided by its molar mass.

Mathematically,   No. of moles = \frac{\text{mass}}{\text{molar mass}}

Hence, we will calculate the moles of oxygen as follows.

       No. of moles = \frac{\text{mass}}{\text{molar mass}}

     Moles of O_{2} = \frac{20.0 g}{32 g/mol}

                            = 0.625 moles

Now,   moles of CO_{2} = \frac{4.4 g}{44 g/mol}

                                      = 0.1 moles

Therefore, total number of moles present are as follows.

Total moles = moles of O_{2} + moles of CO_{2}

                    = 0.625 + 0.1

                    = 0.725 moles

And, total temperature  will be:

                    T = (21 + 273) K = 294 K

According to ideal gas equation,  

                         PV = nRT

Now, putting the given values into the above formula as follows.

                P = \frac{nRT}{V}

                   = \frac{0.725 mol \times 0.08206 Latm/mol K \times 294 K}{5.00 L}

                    = \frac{17.491089}{5} atm

                    = 3.498 atm

or,                = 3.50 atm (approx)

Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.

4 0
3 years ago
What is measured by the heat of reaction?
bixtya [17]

Answer:

The heat of reaction is basically the energy that is being released and or absorbed when chemicals are transformed in a type of chemical reaction. However, the heat of reaction or also called reaction enthalpy is mostly or typically expressed as a molar enthalpy in kJ/mol and or as just a specific enthalpy in kJ/kg or kJ/L.

6 0
3 years ago
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