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3 years ago

HELPPP LOOK AT PICTURE

Chemistry

1 answer:

lina2011 [118]3 years ago

5 0

Answer:

it is a square

Explanation:

I hope it will helps you

sorry it's not a true answer

because I want points

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How many molecules of HCO3 are in 4 moles of the substance?

never [62]

This question is about Avogadro's number.

Avogadro's number is the number is 6.022 * 10^23.

That means that 1 mol of molecules is 6.022 * 10^23 molecules, so 4 moles is 4 * 6.022 * 10^23 molecules = 24.088 * 10^23 molecules = 2.4088 * 10^24 molecules.

Answer: 2.4088 * 10^24 molecules.

8 0

3 years ago

Someone help me please i’ll give brainly

quester [9]

Answer:

Explanation:

i looked it up on Google

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3 years ago

The diagram below shows the PH values of several substances.

skelet666 [1.2K]

Answer:

Changes in colour of litmus paper.

Blue litmus turns red under acidic conditions.
Red litmus turns blue under basic conditions.

Noe

PH values:-

K=4
M=11

K is acidic as pH is <7

Hence K will change the colour of blue litmus paper.

Examples of substances

K=Vinegar, Tomatoes.
M=Milk of magnesia,Soap

4 0

3 years ago

There is an equal number of protons and __________ in a neutral atom.

tiny-mole [99]

the answer is electrons

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3 years ago

What is the percent composition of N H S O in (NH4)2SO4

Lena [83]

Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

- 2 moles of N.

- 8 moles of H.

- 1 mole of S.

- 4 moles of O.

2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

- 2 moles of N (14g/mol) = 28g

- 8 moles of H (1g/mol) = 8g

- 1 mole of S (32g/mol) = 32g

- 4 moles of O (16g/mol) = 64g

3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

$\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}$ $\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}$ $\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}$ $\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\ \\ \end{gathered}$

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

8 0

1 year ago