All categories

3 years ago

If the density of water is 1g/ml, what is the mass of 254ml of water?

1 answer:

8 0

The mass of 254 mL of water is 254 g. Since the density of water is 1g/mL, we can simply multiply the density 1g/mL by 254 mL of water and get 254 g as our answer. Since mL is in the numerator and denominator, mL cancels out and we are left with g only.

You might be interested in

The distinctive odor of vinegar is due to acetic acid, HC₂H₃O₂, which reacts with sodium hydroxide. If a 3.45 mL sample of vineg

Vilka [71]

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH

= 48.875 x 10⁻⁴ moles NaOH

It will react with same number of moles of acetic acid

So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴

number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles

= 1.4167 moles

= 1.4167 x 60 gram

= 85 grams .

So 85 grams of acetic acid will be contained in one litre of acetic acid.

8 0

3 years ago

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago

Please fill this formula

Maurinko [17]

1.) Na
2.) Cl ( at the second blank)
sodium metal+hydrochloric acid

3 0

3 years ago

Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W

nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

H2= (1-x)Haq+XHvap.........1

Putting the values in 1

2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

x = 1904.976 (J/g)/1998.576 (J/g)

x = 0.953

So, the quality of the wet steam is 0.953

7 0

3 years ago

Read 2 more answers

Consider the structures of vinegar and triglyceride and draw in any bond dipoles that exist in the molecules above using a dipol

bagirrra123 [75]

Answer:

Given molecules are vinegar and triglycerides.

Explanation:

The dipole is a vector quantity and it is heading from less electronegative atom to more electronegative atom in a polar covalent bond.

The structures and the bond dipoles in the given molecules are shown below:

5 0

3 years ago