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2 years ago

12

A student viewed the moon through binoculars one week after a new moon.Which image shows the phases of the moon that the student

s observed?

2 answers:

Igoryamba2 years ago

6 0

Answer:

Hi! I think it's maybe H...

Karolina [17]2 years ago

6 0

Answer:The answer is D

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In order to predict the outcome of the reaction, write the molecular, full ionic, and net ionic equations for a mixture of aqueo

Elina [12.6K]

Answer:

See explanation

Explanation:

Full molecular equation;

2NH3(aq) + AgNO3(aq) -------> [Ag(NH3)2]NO3(aq)

Full ionic equation

2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)

Net ionic equation;

2NH3(aq) + Ag^+(aq) --------> [Ag(NH3)2]^+(aq)

When Silver nitrate is mixed with a solution of aqueous ammonia, a white and cloudy solution was observed.

6 0

3 years ago

You carefully weigh out 11.00 g of caco3 powder and add it to 44.55 g of hcl solution. you notice bubbles as a reaction takes pl

Zinaida [17]

The bubbles that were observed after the mixing of the two substances is one of the products of the reaction. It is the carbon dioxide that is produced. To determine the mass of this gas produced, we need to remember the Law of conservation of mass where mass cannot be created or destroyed. With this, we can say that the total mass that goes in a process should be equal to the mass that is goes out of the process no matter what the reaction is. We do as follows:

Mass of reactants = mass of products
11.00 + 44.55 = 51.04 + mass of carbon dioxide
mass of carbon dioxide = 4.51 g

8 0

3 years ago

How would you compare a health care system in Ancient Egypt and a interdisciplinary health care team?

dezoksy [38]

Answer:

A

Explanation:

7 0

2 years ago

A student used 1.168g of an unknown weak acid and titrated it against 28.75 mL of 0.105M NaOH to reach the equivalence point. Wh

Vaselesa [24]

Answer:

The molar mass of the unknown acid is 386.8 g/mol

Explanation:

Step 1: Data given

Mass of the weak acid = 1.168 grams

volume of NaOH = 28.75 mL = 0.02875 L

Molarity of NaOH = 0.105 M

Since we only know 1 equivalence point, we suppose the acid is monoprotic

Step 2: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.105 M * 0.02875 L

Moles NaOH = 0.00302 moles

We need 0.00302 moles of weak acid to neutralize the NaOH

Step 3: Calculate molar mass of weak acid

Molar mass = mass / moles

Molar mass = 1.168 grams / 0.00302 moles

Molar mass = 386.8 g/mol

The molar mass of the unknown acid is 386.8 g/mol

3 0

3 years ago

At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +

Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0

2 years ago