In which types of cells must mutations occur in order for the mutation to be passed on to future generations?

A student needs exactly 1 gram of baking powder to conduct an experiment. The balance reads 0.37 grams. How

sammy [17]

Answer:

B

Explanation:

If the student needs one gram but so far only has 0.37 grams, then the amount they need is the difference between what they need and how much they already have. 1-0.37=0.63 grams.

...which isn't actually an option because none of them have decimal points but I would say it is B anyway because it is the equivalent ratio and maybe there was a typo.

Hope this helped!

6 0

3 years ago

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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

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

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

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

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

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

2 years ago

Which statement describes the phase change that occurs when dry ice is placed in an open container at room temperature?

zloy xaker [14]

Answer:

Dry ice undergoes sublimation, an endothermic change at room temperature.

7 0

3 years ago

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Evidence that best supports the theory of biological evolution was obtained from the

Alisiya [41]

The homologous structures and the analogous structures of different species.

6 0

2 years ago

Find the normality of 0.321 g sodium carbonate in a 250 mL solution.

Radda [10]

Answer:

the normality of the given solution is 0.0755 N

Explanation:

The computation of the normality of the given solution is shown below:

Here we have to realize the two sodiums ions per carbonate ion i.e.

N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)

= 0.1886eq ÷ 0.2500L

= 0.0755 N

Hence, the normality of the given solution is 0.0755 N

5 0

2 years ago