A. How many moles AgNO3 will need to make 6.5 moles of Ca (NO3)2?

How many liters of 1.2M solution can be prepared with 0.50 moles of C6H12O6

BigorU [14]

First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L

5 0

3 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

You would be most likely to use a slicing machine if you were using the __________ method to produce cookies.

Verizon [17]

You would be most likely to use a slicing machine if you were using the icebox method to produce cookies.

In the icebox method a type of cookie in which the dough is made, rolled into a stick, and refrigerated until the dough hardens. The dough can be removed from the refrigerator, cut into individual pieces, and then baked. The rest of the dough is returned to the refrigerator until needed.

Icebox method, also known as refrigerator cookies, are sliced and baked cookies. The dough is formed into logs, chilled in the refrigerator (also called an icebox), sliced , and then baked.

Learn more about the Icebox method here,

https://brainly.in/question/1513677

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7 0

1 year ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

Which sentence best describes an igneous intrusion

Ulleksa [173]

What are the sentences?
It would help to see the sentences.

5 0

3 years ago

Read 2 more answers