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Rasek [7]
3 years ago
9

Why is Iron found as an Ore in the Earth

Chemistry
1 answer:
Stella [2.4K]3 years ago
7 0

\huge \orange {\mathbb {Answer}}

Ore is a deposit in Earth's crust of one or more valuable minerals. The most valuable ore deposits contain metals crucial to industry and trade, like copper, gold, and iron. Copper ore is mined for a variety of industrial uses. Copper, an excellent conductor of electricity, is used as electrical wire.

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58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2,
erica [24]

<em><u>Answer and Explanation:</u></em>

Greetings!

Let's~answer~your~question!

Partial ~pressure ~of ~gas ~can ~be ~directly~ calculated ~by ~multiplying ~the~ percentage\\ of~ pressure~ of~ gases~ to ~the ~total ~pressure.

\boxed{Pgas~ = ~P~total~ * \% ~P ~of ~gas}

<em><u>For % of N2 gas: </u></em>

<em><u /></em>100\% - (5\% + 12\%) = 83\% ~N2<em><u /></em>

<em><u /></em>

<em><u /></em>PN_2 ~= ~146~ atm~ *~ 0.83 ~ = 121.18 ~atm<em><u /></em>

<em><u /></em>PO_2 ~= ~146~ atm~ *~ 0.12~ = 17.52 ~atm<em><u /></em>

<em><u /></em>~PCO_2~ = ~146~ atm ~* 0.05 = 7.3 ~atm<em><u /></em>

3 0
3 years ago
Every atom has atleast two shells ?
VladimirAG [237]
Nope. Hydrogen has one.
4 0
3 years ago
Read 2 more answers
What information would you find on a WHIMIS label?
Nady [450]
Answer/explanation:

Product identifier – the brand name, chemical name, common name, generic name or trade name of the hazardous product.
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8 0
2 years ago
what is the differential rate expression of the change in B with time? Incorporate elements that describe both its formation and
Ivanshal [37]
The differential rate expression for the rate of change in the concentration of B with time is
-rB = dCB/dt = kCB^n
where k is the rate constant and
n is the order of the reaction
This is assuming that the rate is only affected by the concentration of B and the order of the reaction is in the nth order.
5 0
3 years ago
An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu
chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

The density of a proton is 6.278\times 10^{14} g/cm^3.

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\pi r^3..[1]

Diameter of the nucleus ,d' = 9.00\times 10^{-5}\AA

Radius of the nucleus ,r' = 0.5 d'=0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA

Volume of nucleus = V'

V=\frac{4}{3}\pi r'^3..[2]

Dividing [2] by [1]

\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}

=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}

\frac{V'}{V}=4.6656\times 10^{-14}

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

Diameter of the proton ,d = 1.72\times 10^{-15} m = 1.72\times 10^{-13} cm

1 m = 100 cm

Radius of the proton,r = 0.5 d=0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3

Mass of proton, m = 1.0073 amu = 1.0073\times 1.66054\times 10^{-24} g

1 amu = 1.66054\times 10^{-24} g

Density of the proton : d

d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3

The density of a proton is 6.278\times 10^{14} g/cm^3.

5 0
3 years ago
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