A boy lifts a 30N dragon 2 meters above the ground. How much work did the boy do on the dragon?

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0

2 years ago

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2

VLD [36.1K]

Answer:- 64015 J

Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.

density of water is 1 g per mL.

So, the mass of water = $4250mL(\frac{1g}{1mL})$ = 4250 g

Final temperature of water after adding the hot copper bar to it is 26.15 degree C.

So, $\Delta T$ for water = 26.15 - 22.55 = 3.60 degree C

Specific heat for water is 4.184 $\frac{J}{g.^0C}$

The heat gained by water is calculated by using the formula:

$q=mc\Delta T$

where, q is the heat energy, m is mass and c is specific heat.

Let's plug in the values in the formula and do the calculations:

$q=4250g*\frac{4.184J}{g.^0C}*3.60^0C$

q = 64015 J

So, 64015 J of heat is gained by the water.

5 0

2 years ago

Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample

Dimas [21]

Answer:

3.27·10²³ atoms of O

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05 $\frac{g}{mol}$ .

We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.

19.3g Na₂SO₄ · $\frac{1 mol Na^2SO^4}{142.05g Na^2SO^4}$ · $\frac{4 mol O}{1 mol Na^2SO^4}$ · $\frac{6.022x10^2^3}{1 mol O}$

After doing the math for this dimensional analysis, you should get a quantity of approximately 3.27·10²³ atoms of O.

3 0

2 years ago

Which of the following contains maximum number of atoms? a) 2.0g hydrogen b) 2.0g oxygen c) 2.0g nitrogen d) 2.0g methane

Fynjy0 [20]

A) 2.0g hydrogen is your answer

7 0

2 years ago

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered

Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

5 0

1 year ago