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3 years ago

a 32 gram radioactive sample has a half-life of 2 days how much of the original sample remains after 6 days

Chemistry

1 answer:

DiKsa [7]3 years ago

4 0

<span>After 6 days, only 4 grams of the original amount will remain after 6 days. I have attached an image on how to calculate half-life so you can plug your give value into the formula.

</span><span>
To add, </span>half-life means for<span> a substance decaying exponentially, the amount of time it takes for the amount of the substance to diminish by </span>half.

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What is the concentration of a solution made by diluting 25ml of 6.0M HCI to a final volume of 750

Varvara68 [4.7K]

Answer:

Explanation:

The

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3 years ago

A textbook measures 250 mm long, 225 mm wide and 50 mm thick. What is the volume of this book in mm3? What is the volume of this

Dvinal [7]

Answer:

2.81 × 10⁶ mm³

2.81 × 10⁻³ m³

Explanation:

Step 1: Given data

Length (l): 250 mm

Width (w): 225 mm

Thickness (t): 50 mm

Step 2: Calculate the volume of the textbook

The book is a cuboid so we can find its volume (V) using the following expression.

V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³

Step 3: Convert the volume to cubic meters

We will use the relationship 1 m³ = 10⁹ mm³.

2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³

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3 years ago

Let's just say I want to get my mixture to a certain pH but I add too much water to my solution. Can I just add the same volume

Dafna11 [192]

Answer:

You can

Explanation:

The concentration doesn't change when its equals out.

4 0

3 years ago

A car starts from rest. It takes 6.0 seconds for the car to accelerate to 5 m/s2 W. What is the final velocity?

Artyom0805 [142]

Answer: 30 m/s

Explanation:

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3 years ago

formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b

HACTEHA [7]

Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

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Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

3 0

3 years ago