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3 years ago

What activity is a function of all living things?

Chemistry

1 answer:

yaroslaw [1]3 years ago

7 0

I think is cells is the function

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Which information is represented by a line on this map?

Ivanshal [37]

The correct answer is: [A]: "<span>points with the same elevation" .
________________________________________________________</span>

4 0

3 years ago

Nuclear decay<br>27Al + He 》 39P<br>

dybincka [34]

Answer:

$_{13}^{27}\text{Al} + \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

Explanation:

The unbalanced nuclear equation is

$^{27}\text{Al} + \rm \text{He} \longrightarrow \, ^{31}\text{P}$

We can insert the subscripts, because these are the atomic numbers of the elements

$_{13}^{27}\text{Al} + \, \rm _{2}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

That leaves only the superscript of He to be determined,

The main point to remember in balancing nuclear equations is that the sums of the superscripts must be the same on each side of the equation.

Then

27 + x = 31, so x = 31 - 27 = 4

Then, your nuclear equation becomes

$_{13}^{27}\text{Al} + \, \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

6 0

3 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1

H = $\frac{0.1298}{0.0652}$ = 2

O = $\frac{0.130}{0.0652}$ = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is $CH_2O_2$ .

3 0

3 years ago

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the

Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and $NH_3$ .

$\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles$

$\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $NH_3$

So, 0.243 mole of $NaCl$ react with 0.243 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $NaCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaHCO_3$