Which molecule is butyne?

What is the half-life of a 12 g sample of radioisotope that decayed to 6 g in 28

Charra [1.4K]

Answer:

A. 28 years

Explanation:

Applying,

R = R'(2ᵃ/ⁿ).............. Equation 1

Where R = Original sample, R' = Sample left after decay, a = Total time taken to decay, n = half life.

From the question,

Given: R = 12 g, R' = 6 g, a = 28 years.

Substitute into equation 1 and solve for n

12 = 6(2²⁸/ⁿ)

12/6 = 2²⁸/ⁿ

2²⁸/ⁿ = 2

Equation the base,

28/n = 1

n = 28 years.

Hence the half-life is 28 years

4 0

3 years ago

The population of the Earth at the end of 2018 was approximately 7.7 × 10 9 people. Which of the following is the correct way of

stepladder [879]

Answer:

7700000000

Explanation:

so 7.7 x 10 to the power of 9 is in standard form

to put it into a number you just need to times it by 1000000000

however your answer will only have 8 zeros because of the decimal point

5 0

3 years ago

Which type of erosion most likely formed the Grand Canyon?

butalik [34]

Flowing water of course

4 0

3 years ago

The rapid movement of gases molecules can be explained because gases have ___ size particles and exert _____ attraction for othe

Marrrta [24]

The answers that fit the blanks are SMALL and LITTLE, respectively. The particles or molecules or fas are small which makes it loose and easily moves around, and these only exert little attraction for other gas particles. The answer for this would be option D.

5 0

3 years ago

Read 2 more answers

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago