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3 years ago

What are millimeters

2 answers:

6 0

The millimetre or millimeter is a unit of length in the metric system, equal to one thousandth of a metre, which is the SI base unit of length. Therefore, there are one thousand millimetres in a metre. There are ten millimetres in a centimetre.

DerKrebs [107]3 years ago

3 0

Millimeters are an SI unit of length that =0.0001 m

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PLEASE HELP WILL AMRK BRAINLIEST<br> what is the atomic mass of Na3PO4?

Ira Lisetskai [31]

Answer:

163.94

Explanation:

Na = 22.99 (3)

P = 30.97

O = 16.00 (4)

22.99(3) + 30.97 +16.00(4) = 163.94

Hope that helps

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2 years ago

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cupoosta [38]

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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

A bowling ball (mass = 7.2 kg, radius = 0.10 m) and a billiard

evablogger [386]

Answer:

Maximum gravitational Force: $F_{gmax} = 1,026*10^{-08} N$

Explanation:

The maximum gravitational force is achieved when the center of gravity are the closer they can be. For the spheres the center of gravity is at the center of it, so the closer this two centers of gravity can be is:

bowling ball radius + billiard ball radius = 0,128 m

The general equation for the magnitude of gravitational force is:

$F_{gmax} = G \frac{M*m}{r_{min}^{2} }$