CHEMISTRY MIDDLE SCHOOL

Metalloids are border of metals and non metals why​

Answers

Answer 1
Answer:

Answer:

Metalloids are called the border of metals and non metals.

Explanation:

They show some properties of metals and some properties of non metals.
Answer 2
Answer:

Answer:

Metalloids exhibit metal and non-metal properties.

Explanation:

Some metals such as Boron and silicons and not necessarily a metal, nor are they a gas.

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3. For the following reaction, calculate how many grams of each product are formed when 4.05 gof water is used.
2 H2O → 2 H2 + O2

Answers

Answer:

\large \boxed{\text{0.453 g of H$_(2)$; 3.60 g of O$_(2)$}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

MM:                    2.016    32.00

            2H₂O ⟶ 2H₂   +   O₂

m/g:      4.05

1. Mass of hydrogen

(a) Moles of H₂O

\text{Moles of H$_(2)$O} = \text{4.05 g H$_(2)$O}* \frac{\text{1 mol H$_(2)$O}}{\text{18.02 g H$_(2)$O }}= \text{0.2248 mol H$_(2)$O}

(b) Moles of H₂

\text{Moles of H$_(2)$} = \text{0.2248 mol H$_(2)$O } * \frac{\text{2 mol H$_(2)$}}{\text{2 mol H$_(2)$O }} = \text{0.2248 mol H$_(2)$}

(c) Mass of H₂

\text{Mass of H$_(2)$} =\text{0.2248 mol H$_(2)$} * \frac{\text{2.016 g H$_(2)$}}{\text{1 mol H$_(2)$}} = \textbf{0.453 g H$_(2)$}\n\n\text{The reaction produces $\large \boxed{\textbf{0.453 g}}$ of H$_(2)$}

2. Mass of oxygen

(a) Moles of O₂

\text{Moles of O$_(2)$} = \text{0.2248 mol H$_(2)$O } * \frac{\text{1 mol O$_(2)$}}{\text{2 mol H$_(2)$O}} = \text{0.1124 mol O$_(2)$}

(b) Mass of O₂

\text{Mass of O$_(2)$} =\text{0.1124 mol O$_(2)$} * \frac{\text{32.00 g O$_(2)$}}{\text{1 mol O$_(2)$}} = \textbf{3.60 g O$_(2)$}\n\n\text{The reaction produces $\large \boxed{\textbf{0.453 g of H$_(2)$ and 3.60 g of O$_(2)$}}$}

What determines the life cycle of a star

Answers

Hi There!

What determines the life cycle of a star?

A star's life cycle is determined by its mass! :)
A star's life cycle is determined by its mass. The larger its mass, the shorter its life cycle. A star's mass is determined by the amount of matter that is available in its nebula, the giant cloud of gas and dust from which it was born

Consider the following generic reaction: A+2B→C+3D, with ΔH = 161 kJ . Determine the value of ΔH for each of the following related reactions.3A+6B→3C+9D

Answers

Answer:

ΔH=483 kJ

Explanation:

Its given that

For the reaction A+2B→C+3D has  ΔH = 161 kJ.

As the ΔH is positive this is an endothermic reaction where we supply energy.

It is given that we supposed to find ΔH for the following equation.

3A+6B→3C+9D    ΔH=?

So we observe that

3*(A+2B)→3*(C+3D)

So the whole reaction is rised with a coefficient of 3 hence the ΔH would be three times the original one.

ΔH=3*161 kJ

ΔH=483 kJ

Final answer:

The enthalpy change (ΔH) for the reaction 3A+6B→3C+9D is three times the original ΔH, given that the original reaction A+2B→C+3D has a ΔH of 161 kJ. Thus, ΔH for the related reaction is 483 kJ.

Explanation:

The reaction given: A+2B→C+3D, has an enthalpy change (ΔH) of 161 kJ per stoichiometric coefficient of the reaction. Therefore, when you multiply the reaction by three to give 3A+6B→3C+9D, the ΔH value will also be multiplied by three. Thus, the ΔH for the reaction 3A+6B→3C+9D is 161 kJ x 3, which is equal to 483 kJ.

This is because ΔH for a reaction is directly proportional to the number of moles or the quantity of reactants and products involved as shown by Hess's law. For instance, if you have twice or three times the quantity of reactants, you would release or absorb twice or three times that amount of energy, respectively.

Moreover, the enthalpy change of a reaction can also be determined by the enthalpies of formation of the reactants and products or the bond energies depending on the details available. But in this case, the information given directly allows for a simple multiplication to find the answer.

Learn more about Enthalpy Change here:

brainly.com/question/32882904

#SPJ3

A rigid canister holds 50.0 L of helium gas at 32.0°C and 49.9 kPa. The gas is cooled until the pressure reaches 14.2 kPa. What is the Kelvin temperature of the helium?

Answers

Answer:-

86.794 K

Explanation:-

Initial temperature T 1 = 32 C + 273 = 305 K

Initial Pressure P 1 = 49.9 KPa

New Pressure P 2 = 14.2 KPa

The Volume remains same since it is rigid.

∴V 1 = V 2 = 50.0 L

Using the relation

P 1 V 1 / T1 = P 2 V 2 / T 2

New temperature T 2 = P 2 V 2 T 1 / ( P 1 V 1)

= 14.2 KPa x 50.0 L x 305 K / ( 49.9 KPa x 50.0 L)

= 86.794 K

What equation allows you to calculate the force on an object

Answers

Force = mass x acceleration, F = m x a, also F = (m x v)/t
Force = mass times acceleration 

That above is how to calculate force. 

what happens when you mix muriatic acid and phenolphthalein indicator together

Need Help

Answers

When you mix muriatic acid and phenolphthalein indicator together, the solution is colorless. Phenolphthalein indicator will turn pink to basic compounds only not acidic solutions.
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