Please help me i don’t know the answer to this!!!!!!

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Answer 1

Answer:

Answer: Lithium and Ethanol

Explanation: As lithium and ethanol both have a density of under

1.000g/ml, they can be supported by the water. Blood and Magnesium are denser than the water and will therefore sink as the water can not support it them.

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Consider the combination reaction of samarium metal and oxygen gas. If you start with 33.7 moles of samarium metal, how many moles of oxygen gas would be required to react completely with all of the samarium metal? For this reaction, samarium has a +3 oxidation state within the samarium/oxygen compound.

Answers

Answer:

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

$4Sm+3O_2\rightarrow 2Sm_2O_3$

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

$(3)/(4)* 33.7 mol=25.275 mol$ of oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Answer:

Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is $\fbox{25.3 \text{ mol}}$ .

Explanation:

A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.

The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:

1. Amount of one reactant required to react completely with the other reactant.

2. Amount of the product that can be produced from the given amount of the reactant.

Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.

The chemical formula for oxygen gas is $\text{O}_(2)$ .

Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is $\text{Sm}_(2)\text{O}_(3)$ .

The chemical equation is as follows:

$\fbox{\text{Sm}+\text{O}_(2) \rightarrow \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.

The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of $\text{Sm}_(2)\text{O}_(3)$ and 3 in front of $\text{O}_(2)$ to balance the oxygen atoms.

$\fbox{\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.

$\fbox{\n4\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.

According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.

Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:

$\text{moles of O}_(2) = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)$ ...... (1)

Step 5: Substitute 33.7 mol for moles of Sm in equation (1).

$\text{Moles of O}_(2) = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)\n\text{Moles of O}_(2)= 25.275 \text{ mol}\n\text{Moles of O}_(2)= 25.3 \text{ mol}$

Note:

Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Some basic concept of chemistry

Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.

According to kinetci moleculartheory, which of the following would not be onsidered an ideal gas

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Answer:

hope it will help u

thank you

What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for water is 1.86 °C/m.

Answers

Answer:

freezing point (°C) of the solution = - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3

$K_f$ = 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

$molality = (mole \ of \ solute )/(kg \ of \ solvent )$

which can now be re-written as :

$molality = (mole \ of \ Ca(NO_3)_2)/(kg \ of \ water)$

$molality = ((mass \ of \ \ Ca(NO_3)_2)/(molar \ mass of \ Ca(NO_3)_2) )/(kg \ of \ water)$

$molality = ((11.3 \ g )/(164 \ g/mol) )/(0.115 \ kg )$

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

$\Delta T =3 * (1.86 \ ^0C/m) * (0.599 \ m)$

$\Delta T =3.34^0 \ C$

$\Delta T =$ the freezing point of water - freezing point of the solution

3.34° C = 0° C - freezing point of the solution

freezing point (°C) of the solution = 0° C - 3.34° C

freezing point (°C) of the solution = - 3.34° C

Calculate the mass % of magnesium sulfate (assume that there is a 1:1 mol ratio between sulfate and magnesium sulfate) in the original sample. Report your answer without units and use 3 sig figs, i.e. 55.23543% would be entered as 55.2

Answers

61.8 % is the mass percentage of magnesium sulphate.

Explanation:

The mass percent of individual solute or ion in a compound is calculated by the formula:

Grams of solute ÷ grams of solute + solvent × 100

mass percent of magnesium is calculated as 1 mole of magnesium having 24.305 grams/mole will have weight of 24.305 grams and 1 mole of MgSO4 will have 120.366 grams

Putting the values in the equation:

24.305 ÷ 144.671 × 100

= 16.8% of magnesium is in the mixture

The mass percentage of SO4 is calculated as

= 96.06 ÷ 216.426 × 100

= 44.38 %

The mass percentage of the mixture MgSO4 is 44.38 + 16.8 = 61.8 %

Mass percentage is a representation of the concentration of element or elements in a compound.

How would you explain the path of alpha particles in relation to each subatomic particle in the plum pudding model?

Answers

In the plum pudding model, alpha particles were expected to pass through the foil at different angles and some of them reflected. They realized that there are positive charge atom because it repels. So the plum pudding was replaced by the nuclear model of atom.

The length of the marathon race is approximately 26.2 mi. What is the distance in kilometers?