NaCl+F2=NaF+Cl2 balancing chemical equations

Uranium (VIII) Sulfide formula

tresset_1 [31]

Answer:

US₂

Explanation:

Uranium sulfide (US₂)

Uranium atomic symbol = U

Sulfur atomic symbol = S

Uranium valency = +4

Sulfur valency = -2

So;

Uranium sulfide (US₂)

8 0

3 years ago

a soft drink contains 47.0 carbohydrates per serving. calculate the number of nutritional calories (cal) per serving in this dri

Zolol [24]

A soft drink that contains 47 grams of carbohydrates per serving has a nutritional calories (cal) per serving of 188 calories.

Calories are a measurement of energy released from your body after digestion and absorption of food whereas carbohydrate is a carbon-based organic compound consisting of carbon, hydrogen, and oxygen atoms with a general chemical molecular formula of CH₂O.

In order to get the number of nutritional calories per serving from a soft drink that has 47 grams of carbohydrate, we will need to multiply the number of grams by 4 since there are about 4 calories for every gram of carbohydrate, and the calculation is as follows:

47×4=188

Hence, there are 188 calories in a soft drink that contains 47 grams of carbohydrates.

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8 0

1 year ago

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

Give the number of particles of each element in Na2SO4, sodium sulfate, a drying agent.

aleksklad [387]

Sodium, Sulphur, and Oxygen

4 0

3 years ago

Some plants can reproduce both sexually and asexually. Explain one benefit of this dual reproduction capability.

pantera1 [17]

Answer:

An additional advantage of asexual reproduction is that colonization of new habitats may be easier when an individual does not need to find a mate to reproduce. During sexual reproduction the genetic material of two individuals is combined to produce genetically diverse offspring that differ from their parents.

Explanation:

An organism has many benefits for having the ability to reproduce both sexually and asexually. The first of these benefits is that it doesn't have to find a mate in order to reproduce, so it can create offspring by itself, and enable the continuation of its' species.

If you need more help on this subject, don't be afraid to ask me. I'm willing to help.

8 0

3 years ago