In this periodic table, which type of element is shown in green boxes?

According to the law of superposition, where would you find objects with an older relative age?

Advocard [28]

Answer: if you think about a tree the older rings would be in the middle, so you’d find objects with older age below/under/inside

7 0

3 years ago

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Please anyone answer this question

Ainat [17]

Answer:

A, 18.4 and 34!

Explanation:

The Burette seems to have ~18.4 of the liquid? substance in it, and the Measuring Cylinder seems to have ~34 of the liquid? substance in it. So, it is A! :]

Hope this helps!!

7 0

3 years ago

For the aqueous reaction dihydroxyacetone phosphate is the reactant and glyceraldehyde 3 phosphate is the product. dihydroxyacet

lbvjy [14]

Answer: The Gibbs free energy of the reaction is -445 J/mol.

Explanation:

The chemical equation for the conversion follows:

$\text{Dihydroxyacetone phosphate}\rightleftharpoons \text{Glyceraldehyde-3-phosphate}$

The expression for $K_{eq}$ of above equation is:

$K_{eq}=\frac{\text{[Glyceraldehyde-3-phosphate]}}{\text{[Dihydroxyacetone phosphate]}}$

We are given:

[Glyceraldehyde-3-phosphate] = 0.00400 M

[Dihydroxyacetone phosphate] = 0.100 M

Putting values in above equation, we get:

$K_{eq}=\frac{0.004}{0.100}=0.04$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G=\Delta G^o+RT\ln K_1$

where,

$\Delta G^o$ = Standard Gibbs free energy = 7.53 kJ/mol = 7530 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

Putting values in above equation, we get:

$\Delta G=7530J/mol+(8.3145J/Kmol)\times 298K\times \ln (0.04)\\\\\Delta G=-445J/mol$

Hence, the Gibbs free energy of the reaction is -445 J/mol.

3 0

4 years ago

Aluminum oxide is a covalent compound. True False

JulijaS [17]

This statement is false.

Hope this helps,

Davinia. (:

8 0

3 years ago

Read 2 more answers

Percentage Yield

scoundrel [369]

Q.No. 1:

You have 20.0 g of CaCO₃. You decompose it by heat, and weigh the calcium oxide that remains. You have 10.3 grams. What is the % yield?

Answer:

%age Yield = 92.37 %

Solution:

The balance chemical equation for given decomposition reaction is;

CaCO₃ → CaO + CO₂

Step 1: Calculate Moles of CaCO₃:

Moles = Mass / M.Mass

Moles = 20.0 g / 100.08 g/mol

Moles = 0.199 moles of CaCO₃

Step 2: Calculate theoretical amount of CaO produced;

According to equation,

1 moles of CaCO₃ produced = 1 mole of CaO

So,

0.199 moles of CaCO₃ will produce = X moles of CaO

Solving for X,

X = 0.199 mol × 1 mol / 1 mol

X = 0.199 mol of CaO

Also,

Mass = Moles × M.Mass

Mass = 0.199 mol × 56.07 g/mol

Mass = 11.15 g of CaO

Step 3: Calculate %age Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

%age Yield = 10.3 g / 11.15 g × 100

%age Yield = 92.37 %

___________________________________________

Q.No. 2:

A student makes sodium chloride by buming 2.3 grams of sodium in chlorine gas. If the yield is 90%, how much sodium chloride is made?

Answer:

Actual Yield = 5.785 g of NaCl

Solution:

The balance chemical equation for given decomposition reaction is;

2 Na + Cl₂ → 2 NaCl

Step 1: Calculate Moles of Na:

Moles = Mass / M.Mass

Moles = 2.3 g / 23 g/mol

Moles = 0.10 moles of Na

Step 2: Calculate theoretical amount of NaCl produced;

According to equation,

2 moles of Na produced = 2 moles of NaCl

So,

0.10 moles of Na will produce = X moles of NaCl

Solving for X,

X = 0.10 mol × 2 mol / 2 mol

X = 0.10 mol of NaCl

Also,

Mass = Moles × M.Mass

Mass = 0.10 mol × 58.44 g/mol

Mass = 5.844 g of NaCl

Step 3: Calculate Actual Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

Or,

Actual Yield = %age Yield × Theoretical Yield ÷ 100

Actual Yield = 99 × 5.844 g ÷ 100

Actual Yield = 5.785 g of NaCl

___________________________________________

Q.No. 3:

For the chemical reaction Mg(s) + 2 HCl (aq) → H₂ (g) + MgCl₂ (aq) calculate the % yield if 100 grams of magnesium react with excess HCl to produce 310 grams of MgCl₂.

Answer:

%age Yield = 79.13 %

Solution:

The balance chemical equation for given decomposition reaction is;

Mg + 2 HCl → MgCl₂ + H₂

Step 1: Calculate Moles of Mg:

Moles = Mass / M.Mass

Moles = 100 g / 24.30 g/mol

Moles = 4.11 moles of Mg

Step 2: Calculate theoretical amount of MgCl₂ produced;

According to equation,

1 mole of Mg produced = 1 mole of MgCl₂

So,

4.11 moles of Mg will produce = X moles of MgCl₂

Solving for X,

X = 4.11 mol × 1 mol / 1 mol

X = 4.11 mol of MgCl₂

Also,

Mass = Moles × M.Mass

Mass = 4.11 mol × 95.21 g/mol

Mass = 391.73 g of MgCl₂

Step 3: Calculate %age Yield as;

%age Yield = Actual Yield / Theoretical Yield × 100

%age Yield = 310.0 g / 391.73 g × 100

%age Yield = 79.13 %

5 0

4 years ago