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3 years ago

What is 0.000028474 in scientific notation

Chemistry

2 answers:

Luda [366]3 years ago

8 0

Answer:

2.8474 * 10^-5.

Explanation:

There are 5 digits after the decimal point and up to and including the 2 so the exponent is -5.

antiseptic1488 [7]3 years ago

6 0

Answer:

$2.8474 \: \times \: {10}^{ - 5}$

this the solution

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4Fe +302 + 2Fe2O3 <br> How many grams of Fe2O3 will be produced from 10.0 g of Fe and 7.00 g of O2?

Molodets [167]

Answer:10g of Fe produces 14.37g of Fe2O3

7g of O2 produce 46.84g of Fe2O3

Explanation:

3 0

3 years ago

2. Explain Why would it be difficult to find the volume of a rock using<br> a ruler?

dedylja [7]

Answer:

Most of these rocks are not made up of common geometric shapes

Explanation:

Because most rocks are not made up of common geometric shapes, it would be difficult or impossible to find the volume of a rock using a ruler; there would be no easy way to measure the rock's volume using a ruler

Hope this helped!

8 0

3 years ago

We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower lim

eduard

Answer:

The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are

Ku = 38.252 W/mK

K lower = 0.199 W/mK

Explanation:

As we know

Ku = Vp * Kair + Vmagnesium * K metal

Ku = 0.10 *0.02 + (1-0.25) * 51

Ku = 38.252 W/mK

The lower limit

K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)

K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)

K lower = 0.199 W/mK

8 0

3 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

3 years ago

Write this in a word and skeleton equation:

DiKsa [7]

Answer:

Write this in a word and skeleton equation:

Solid silver chloride and an aqueous solution of nitric acid are produced when a solution of silver nitrate is reacted with a solution of hydrochloric acid.

Explanation:

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3 years ago