What is the concentration of a 1.5L solution that contains 0.030 mol of hydrochloric acid?

eleanor wants to show her class a comparison of melting rates between two different solids when he is applied why would a graph

sashaice [31]

Eleanor wants to show her class a comparison of melting rates between two different solids when heat is applied. A graph would be more suitable than a table to show her data because the she can see the trend line of the effects of melting than the table.

7 0

3 years ago

Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4

kozerog [31]

Answer:

The feed ratio (liters 20% solution/liter 60% solution) is 3,08

Explanation:

In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.

100 kg of 20% solution are 100kg/1,139 kg/L = <em>87,8 L </em>

100kg×20wt% = 20 kg H₂SO₄. In moles:

20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ <em>203,9 mol</em>

The final molarity 4,00M comes from:

$\frac{203,9 moles+ Xmoles}{87,8L + Yliters}$ <em>(1)</em>

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L× $\frac{1,498 kgsolution}{L}$ × $\frac{60 kg H_{2}SO_{4}}{100kgSolution}$ × $\frac{1kmol}{98,08kg H_{2}SO_{4}}$ = <em>0,9164 kmolH₂SO₄ ≡ 916,4 moles</em>

That means:

X/Y = 916,4/100 = 9,164 <em>(2)</em>

Replacing (2) in (1):

Y(liters of 60,0 wt% H₂SO₄) = <em>28,52 L</em>

Thus, feed ratio (liters 20% solution/liter 60% solution):

87,8L/28,52L = <em>3,08</em>

I hope it helps!

8 0

3 years ago

How much energy is required to boil 65 grams of 100°C water<br> And then heat the steam to 150°C?

Artyom0805 [142]

Answer:

13598 J

Explanation:

Q = m × c × ∆T

Where;

Q = amount of energy (J)

m = mass (grams)

c = specific heat capacity

∆T = change in temperature

m = 65g, specific heat capacity of water = 4.184J/g°C, initial temperature= 100°C, final temperature = 150°C

Q = 65 × 4.184 × (150 - 100)

Q = 271.96 × 50

Q = 13598 J

Hence, 13598 J of energy is required to boil 65 grams of 100°C water and then heat the steam to 150°C.

8 0

3 years ago

PLEASE HELP FOR THIS ASSIGNMENT

mars1129 [50]

<h2>Steps:</h2>

Remember that Density = mass/volume, or D = m/v

So firstly, we have to find the volume of the rock. To do this, we need to subtract the volume of water A from the volume of the water B. In this case:

Water A = 30 mL
Water B = 40 mL
40 mL - 30 mL = 10 mL

<u>The volume of the rock is 10 mL.</u>

Now that we have the volume, we can plug that and the density of the rock into the density equation to solve for the mass.

$3.678=\frac{m}{10}$

For this, multiply both sides by 10:

$36.78=m$

<h2>Answer:</h2>

<u>Rounding to the tenths place, the mass of the rock is 36.8 g, or C.</u>

6 0

4 years ago

Explain why the quantum number set (2, 2, -1, -½) is not possible for an electron in an atom.

rosijanka [135]

Answer is: it is not possible, because orbital (azimunthal) quantum number cannot be 2.

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

For principal quantum number n=2:

1) azimuthal quantum number (l) can be l = 0...n-1:

l = 0, 1.

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.

2) magnetic quantum number (ml) can be ml = -l...+l.

ml = -1, 0, +1.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.

3) the spin quantum number (ms), is the spin of the electron.

ms = +1/2, -1/2.

8 0

4 years ago