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4 years ago

In reaction CFCl3 + UV Light -> CFCl2 + Cl, there is only one reactant (CFCl3) and no collision

1 answer:

3 0

The reaction CFCl3 + UV Light -> CFCl2 + Cl does not need another reactant as with CFCl3 because the reaction itself is reactive to light. Note that there are reactions that are sensitive to light to form products and when this type of reaction are not exposed to light, no reaction occurs.

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What are the reactants in a neutralization reaction?

castortr0y [4]

answer an element and a compound

8 0

3 years ago

1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

We have the following data:

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:

$P_0*V_0 = P*V$

$1*1 = P*0.473$

$1 = 0.473\:P$

$0.473\:P = 1$

$P = \dfrac{1}{0.473}$

$\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

The new pressure of the gas is 2.11 atm

___________________________________

$\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

3 0

3 years ago

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

Step 1: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

Step 2: Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

Step 3: Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

3 years ago

Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit

Anastasy [175]

The dissociation equation will be

NH4OH ---> NH4+ + OH-

Initial 0.006 0 0

Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053

Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0

3 years ago

Imagine you are making Kool Aid. Kool Aid comes in packets. The powder inside the packets is made up of very small particles. Th

lina2011 [118]

Answer:

Because the molecules have not been in water so they are not moving around each other

Explanation:

4 0

3 years ago

Read 2 more answers