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Whitepunk [10]
2 years ago
12

If another student joins the game and stands behind the two students

Chemistry
1 answer:
Mandarinka [93]2 years ago
4 0

Answer:

ha djd sidi dj sjejeisns

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If 1.38×1022 atoms of element y have a mass of 1.50 g, what is the identity of y? express your answer as a chemical symbol.
zaharov [31]
1,38×10²² = 0,138×10²³

0,138×10²³ ----- 1,5g
6,02×10²³ ------ X
X = (1.5×6,02×10²³)/0,138×10²³
X = 65,435 g/mol

It's ZINC (Zn)

:•)
4 0
3 years ago
Read 2 more answers
Is there a way to dissolve more sugar into a solution that it should hold at that temperature? Explain how this is done and what
Darya [45]

The answer is: supersaturated solution.

A supersaturated solution contains more of the dissolved substance than could be dissolved by the solvent under normal circumstances.

A way to dissolve more sugar into a solution is heating a solution.

The more heat is added to a system, the more soluble a substance (in this example sugar) becomes.

The solution will become supersaturated if this solution is suddenly cooled at a rate faster than the rate of precipitation.

6 0
3 years ago
1. Complete the following
Marat540 [252]

Answer:

\large \boxed{\text{0.603 mol}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

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

Mᵣ:        180.16

           C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:        18.1

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}

b) Moles of O₂

\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}  

3 0
3 years ago
A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate
Zarrin [17]

Answer: Empirical formula is C_2H_5O

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = \frac{330g}{12g/mol}=27.5moles

Moles of Hydrogen = \frac{69.5g}{1g/mol}=69.5moles

Moles of Oxygen = \frac{220.2g}{16g/mol}=13.76moles

<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2

\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5

\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1

<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = C_2H_5O

7 0
3 years ago
A sample of xenon gas occupies a volume of 5.24 L at 411 K. If the pressure remains constant, at what temperature will this
stiks02 [169]

Answer:

161.57 K

Explanation:

V1/T1 = v2/T2

(5.24 L)/(411 K) = (2.06 L)/T2

T2 = 161.57 K

8 0
2 years ago
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