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2 years ago

8

one ball rolls along a shelf at a steady rate. a second ball rolls off the shelf and gains speed as it falls in a curved path. w

hich must have an unbalanced force acting on it?

1 answer:

Salsk061 [2.6K]2 years ago

6 0

The ball that rolls at a steady rate

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What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium c

olya-2409 [2.1K]

Answer:

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

$pOH=pKb+log\frac{[salt]}{[base]}$

pH = 14- pOH

Let us calculate pOH $pOH=3.43+ (-0.397)=3.03$

$pH=14-pOH=14-3.03=10.97$

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

$[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M$

[base] = [Methylamine]=0.10

After mixing with salt

$[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M$

pKb= -log[Kb]= 3.43

Putting values

pOH = $3.43+log(\frac{[0.0286]}{0.0714}$

4 0

2 years ago

The amount of gas that occupies 36.52 L at 68.0°C and 672 mm Hg is __________ mol.

ASHA 777 [7]

We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:

PV = nRT
(672) (1/760) (36.52) = n (0.08206) ( 68 +273.15)
n = 1.15 mol of gas

Hope this answers the question. Have a nice day.

7 0

2 years ago

How many capsules containing 75mg of Tamiflu could be produced from 155g of star anise.?

jeyben [28]

From the question you will find that:
one capsule of tamiflu is obtained from 2.6 g of star anise.
1 capsule = 2.6 g tamiflu
? capsules = 155 g tamiflu
by cross multiplication = $\frac{(1 x 155)}{2.6}$ = 59 capsules

7 0

2 years ago

Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

2 years ago

Give an example of role of luck in bottenheims research

Veronika [31]

"You may find what you seek, or you might not" Is an excellent quote that might help.

3 0

2 years ago

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