A 35.0 mL solution of Sr(OH)2 is neutralized with 29.5 mL of 0.200 M

Which nuclide has a half-life that is less than one minute?

Elena-2011 [213]

I think a good strategy for these kind of problems is to just choose the atom with the highest mass number. At any rate, the answer actually is (and I guessed it as) (2) Francium-220. It has a half life of about 30 seconds.

6 0

3 years ago

Read 2 more answers

Why Can alloys not be described using chemical formulas?

Reptile [31]

Alloys are homogeneous mixtures, they are not compounds like others and hence donot have formula

6 0

4 years ago

Read 2 more answers

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago

Which of the following statements is true about the relationships between photon energy, wavelength, and frequency?

iris [78.8K]

Answer: The Answer is A.

Explanation:

The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.

Hope this Helps!

8 0

4 years ago

How many Liters of 0.968M solution can be made if 0.581 moles of solute are added? Group of answer choices 0.600 L 60 mL 0.562 L

Rzqust [24]

Answer:

0.6L

Explanation:

The formula of molarity is molSolute/litreSolution

$0.968M=\frac{0.581}{LitreSolution} \\\\LitreSolution=\frac{0.581}{0.968} \\LitreSolution=0.6L$ .

7 0

4 years ago