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3 years ago

1,5,25,125 what’s the pattern rule and extend by 3 more numbers

1 answer:

4 0

For the first one the pattern is multiply the previous number by five as you see 1 x 5 = 5 and so on. To keep adding to it you would do

125 x 5 = 625 625 x 5 = 3125 3125 x 5 = 15625

Now for the second one the pattern is divide the previous number by three as you can see 2187 / 3 = 729 and so on. To keep going you would

81 / 3 = 27 27 / 3 = 9 9 / 3 = 3

I hope this helps you and if you have anymore questions i'll be glad to answer them.

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Which elements from the list would be likely to form an ionic bond with chlorine, Cl?

monitta

Answer:

metal calcium or (Ca).

Explanation:

For example, the metal calcium (Ca) and the nonmetal chlorine (Cl) form the ionic compound calcium chloride (CaCl2). In this compound, there are two negative chloride ions for each positive calcium ion

8 0

3 years ago

he standard enthalpies of formation for S (g), F (g), SF4 (g), and SF6 (g) are +278.8, +79.0, -775, and -1209 kJ per mole, respe

JulsSmile [24]

Answer:

+60.54 kJ/mol

Explanation:

The enthalpy of formation of a compound is the sum of the enthalpies of the formation of its constituents and the bond of them. To form a compound, energy must be lost, so, they're negative.

For SF4, the enthalpy is formed by the energy of one S, two F, and 4 S-F bond (Hb):

H = - (278.8 + 4*79.0 + 4*Hb)

-775 = -(594.8 + 4Hb)

594.8 + 4Hb = 775

4Hb = 180.2

Hb = +45.05 kJ/mol

For SF6, the enthalpy is formed by the energy of one S, six F, and 6 S--F bonds (Hb):

H = -(278.8 + 6*79.0 + 6*Hb)

-1209 = -(752.8 + 6Hb)

752.8 + 6Hb = 1209

6Hb = 456.2

Hb = +76.03 kJ/mol

Thus, the energy of S--F bond must be the average of these two:

(45.05 + 76.03)/2 = +60.54 kJ/mol

3 0

3 years ago

This type of erosion picks up materials and transports them in the air, often causing the abrasion of surfaces.

SVEN [57.7K]

Answer: Sediment Transport by Wind

Explanation: Like flowing water, wind picks up and transports particles. Wind carries particles of different sizes in the same ways that water carries them (Figure below). Tiny particles, such as clay and silt, move by suspension. They hang in the air, sometimes for days.

6 0

3 years ago

What do repeated trials in an experiment allow a scientist to do?

Oksana_A [137]

Answer:

Learn from their experiment and know the possible results from trial and error

8 0

4 years ago

The decay constant for 14C is .00012 In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed

Andru [333]

The question is incomplete, here is the complete question:

The decay constant for 14-C is $0.00012yr^{-1}$ In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.

<u>Answer:</u> The formula for the age of the charcoal is $t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

<u>Explanation:</u>

Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 20) = 80 grams

Putting values in above equation, we get:

$1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

Hence, the formula for the age of the charcoal is $t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

5 0

4 years ago