Density of 2.7 g/ml and volume of 35.6 ml what is the mass

Chemistry

1 answer:

Anettt [7]3 years ago

3 0

The math is set up like

35.6 ml * 2.7 g/ 1 ml

which will leave you with

96.12 g

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A technical machinist is asked to build a cubical steel tank that will hold 195 L of water. Calculate in meters the smallest pos

Aleks04 [339]

Answer:

The smallest possible inside length of the tank is $0.579$ m.

Explanation:

As we know that

$1 m^3 = 1000 L$

Thus, volume of $195$ liter tank is also equal to $0.195$ cubic meter

The volume of a cube is equal to $x^3$ , where, x is the length of the side of the cube

With the give condition,

$x^ 3 = 0.195$

Solving the above equation, we get -

$x = (0.195)^{\frac{1}{3})}\\x = 0.579$

The smallest possible inside length of the tank is $0.579$ m.

4 0

3 years ago

Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The

Rudiy27

<span>
1. Remember (sum of products) - (sum of reactants)
So ΔHrxn = 2 ΔHf [H2(g)] + ΔHf [Ca(OH)2(s)] - 2 ΔHf [H2O(l)] - ΔHf [Ca(s)]
= 2*0 + -986.09 kJ/mol - 2*(-285.8 kJ/mol) - 0

Do the math and you'll have the answer. BTW the ΔHf [H2(g)] and ΔHf [Ca(s)] were 0 because these are elements in their standard states.

</span>HOPE THIS HELPS ;)

8 0

3 years ago

Read 2 more answers

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

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

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$