I NEED HELP PLEASE, THANKS! :) The density of water at 4.00°C is 0.967 g/mL. How many molecules of water are present in a 499.8 mL bottle of water? Express your answer to the correct number of significant figures.

The number of molecules of water  present in the bottle is 1.62×10²⁵ molecules.

We'll begin by calculating the mass of the water in the bottle.

Mass = Density × Volume

Mass of water = 0.967 × 499.8

Mass of water = 483.3066 g

Finally, we shall determine number of molecules of water in the bottle.

From Avogadro's hypothesis,

1 mole of water = 6.02×10²³ molecules

But,

1 mole of water = 18 g

Thus, we can say that:

18 g of water = 6.02×10²³ molecules

Therefore,

483.3066 g of water = (483.3066 × 6.02×10²³) / 18

483.3066 g of water = 1.62×10²⁵ molecules

Thus, the number of molecules of water in the bottle is 1.62×10²⁵ molecules.

Learn more about Avogadro's number:

link

Explanation:

You must calculate the mass of the water, convert it to moles, and then calculate the number of molecules.

1. Mass of water

2. Moles of water

3. Molecules of water

where's the diagram?

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol