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2 years ago

NO BOTS! Determine the total net force and the direction of movement for the situation.

Chemistry

1 answer:

pychu [463]2 years ago

5 0

Answer:

5N Left (the direction of friction)

Explanation:

Looking at the normal force and the gravitational force, they are both of equal magnitude (3N), and are in opposite directions ( up and down). This means they cancel out - their overall force is 0.

Moving on to the sideways forces, we only have friction (5N). There is no force to the right to cancel this out, so the overall force must be 5N left.

Hope this helps! Let me know if you have any questions :)

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CuCl2 + NaNO3 -------> Cu(NO3)2 + NaCl

pav-90 [236]

Answer:

what

Explanation:

3 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

A variation of the acetamidomalonate synthesis can be used to synthesize serine. The process involves the following steps: Ethox

Alex73 [517]

Answer:

See detailed mechanism in the image attached

Explanation:

The mechanism shown in detail below is the synthesis of serine in steps.

The first step is the attack of the ethoxide ion base on the diethyl acetamidomalonate substrate giving the enolate and formaldehyde.

The second step is the protonation of the oxyanion from (1) above to form an alcohol as shown.

Acid hydrolysis of the alcohol formed in (3) above yields a tetrahedral intermediate, a dicarboxyamino alcohol.

Decarboxylation of this dicarboxyamino alcohol yields serine, the final product as shown in the image attached.

5 0

3 years ago

When producing a soluble salt in a reaction between an acid and an alkali, how can you prepare dry solid crystals from the solut

igor_vitrenko [27]

Answer:

aqueous solution

Explanation:

7 0

3 years ago

A laboratory procedure calls for making 590.0 mL of a 1.1 M KNO3 solution. How much KNO3 in grams is needed?

AveGali [126]

590 mL = 590 cm³= 0,59 dm³

C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________

M KNO₃ = 39g+14g+16g×3 = 101 g/mol

1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃

:)

8 0

3 years ago