3. Measure: Find the mass, volume, and density of each of the three crowns.

Wittaler [7]

Answer:

2.475 mol of O2 formed.

Explanation:

Given 1.65 moles of KClO3 as the target amount in the reactant, used the coefficient of the balanced chemical reaction involved to determine the number of moles of O2 molecules formed.

x mole of O2 = 1.65 mol KClO3 x [(3 mol O2)/ (2 mol KClO3)] = 2.475 mol of O2

x mole of O2 formed = 2.475 mol of O2

5 0

3 years ago

A gas is contained in a thick-walled balloon. When the pressure change from 100 kPa to 90.0 kPa, the volume changes from 2.50 L

Orlov [11]

The temperature change is calculated using the combined gas law
that is P1V1/T1 =P2V2/T2
P1= 100KPa
P2=90kpa
v1= 2.50 L
v2= 3.75 L
T1= 303 K
T2=?

T2 is therefore = P2V2T1/P1V1
=( 90 x 3.75 x303)/ (100 x2.50) = 409.05 K

3 0

3 years ago

Read 2 more answers

An instrument used to separate cell parts according to density is the

grigory [225]

It’s a centrifuger there are many steps to this do I advise looking on the internet or i could explain if needed

7 0

4 years ago

Using the appropriate ksp value from appendix d in the textbook, calculate the ph of a saturated solution of ca(oh)2 .

Alex777 [14]

The ph of a saturated solution of Ca(OH)2 is 12.35

CALCULATION:
For the reaction
Ca(OH)2 → Ca2+ + 2OH-
we will use the Ksp expression to solve for the concentration [OH-] and then use the acid base concepts to get the pH:
Ksp = [Ca2+][OH-]^2

The listed Ksp value is 5.5 x 10^-6. Substituting this to the Ksp expression, we have
Ksp = 5.5 x 10^-6 = (s) (2s)^2 = 4s^3
s3 = 5.5x10^-6 / 4

Taking the cube root, we now have
s = cube root of (5.5x10^-6 / 4)s
= 0.01112

We know that the value of [OH-] is actually equal to 2s:
[OH-] = 2s = 2 * 0.01112 = 0.02224 M

We can now calculate for pOH:
pOH = - log [OH-]
= -log(0.02224)
= 1.65

Therefore, the pH is
pH = 14 - pOH
= 14 - 1.65
= 12.35

4 0

4 years ago

How many grams of sodium chromate, Na2CrO4, are needed to react completely with

JulsSmile [24]

Answer:

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

Explanation:

Step 1: data given

Mass of silver nitrate AgNO3 = 56.7 grams

Molar mass AgNO3 = 169.87 g/mol

Molar mass of Na2CrO4 = 161.97 g/mol

Step 2: The balanced equation

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 56.7 grams / 169.87 g/mol

Moles AgNO3 = 0.334 moles

Step 4: Calculate moles Na2CrO4 moles needed

For 2 moles AgNO3 we need 1 mol Na2CrO4 to produce 1 mol Ag2CrO4 and 2 moles NaNO3

For 0.334 moles AgNO3 we need 0.334 / 2 = 0.167 moles Na2CrO4

Step 5: Calculate mass Na2CrO4

Mass Na2CrO4 = 0.167 moles * 161.97 g/mol

Mass Na2CrO4 = 27.0 grams

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

8 0

3 years ago