A wedge, which is used to split objects, distributes the force applied on its blunt face as shown in the image.

Chemistry

2 answers:

nlexa [21]3 years ago

8 0

B.the sides of the wedge are inclined

atroni [7]3 years ago

8 0

Answer:

Explanation:

i did the usatest prep

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Describe what happens when ionic and covalent (molecular) substances dissolve. Fill in the blanks using the options below:covale

Lena [83]

1. ionic compound , aqueous cations and aqueous anions

2. covalent compound aqueous covalent compound

Explanation:

1. A(n) ionic compound dissolves in water , H₂O(l), will produce aqueous cations and aqueous anions in solution.

When NaCl dissolves in water it will produce Na⁺ and Cl⁻ ions in solution

2. A(n) covalent compound  dissolves in water , H₂O(l), will produce aqueous covalent compound in solution.

When Ammonia (NH₃) dissolves in water it forms aqueous ammonia, NH₃(aq)

Organic compounds, like carbohydrates, proteins, nucleic acids, and lipids, are all good examples of covalent compounds.

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3 years ago

When the pressure and number of particles are kept constant for a sample of gas, which of the following is also constant for the

Leno4ka [110]

The answer is D....................

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3 years ago

You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?

umka21 [38]

Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

$N=\frac{eq_{solute}}{V_{solution}}$

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

$eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq$

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

$m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4$