The weak base ionization<br> constant (Kb) for CN is<br> equal to:

I am one of the elements. I am a soft representative metal, and you can cut me with a knife, but watch out because I am highly r

Flauer [41]

Answer:

Potassium

Explanation:

From the given riddle. Potassium is the right answer. This because on the periodic table, potassium has the smallest atomic number of any element in its period with the electronic configuration 1s2 2s2 2p6 3s2 3p6 4s1, it has the largest atomic radius within its group because the radius has no d-electron shell filled. Potassium is a soft metal but contains high reactivity.

5 0

3 years ago

HELP FAST!!!

UNO [17]

the answer is thermal energy

8 0

3 years ago

Read 2 more answers

A copper wire is 44.85 cm long and weighs 1.521 g. The density of copper is 8.933 g/cm3.

PolarNik [594]

Answer:

i am so sorry

Explanation:

7 0

3 years ago

A chemist wishing to do an experiment requiring 47Ca2+(half-life = 4.536 days) needs 5.00 g of the nuclide. What mass of 47CaCO

Artyom0805 [142]

Answer: The answer is 6.78 grams.

Explanation: The equation used for solving this type of problems is:

$\frac{N}{N_0}=(\frac{1}{2})^n$

where, $N_0$ is the initial amount of radioactive substance, N is the remaining amount and n is the number of half lives.

Number of half lives is calculated on dividing the given time by the half life.

n = time/half life

Time is given as 48.0 hours and the half life is given as 4.536 days. let's make the units same and for this let's convert the half life from days to hours.

$4.536days(\frac{24hours}{1day})$

= 108.864 hours

So, $n=\frac{48.0}{108.864}$ = 0.441

Since 5.00 g is the required amount when the radioactive substance is delivered to the scientist, it would be the final amount that is N. We need to calculate the initial amount. Let's plug in the values in the equation:

$\frac{5.00}{N_0}=(\frac{1}{2})^0^.^4^4^1$

$\frac{5.00g}{N_0}=0.737$

$N_0=\frac{5.00g}{0.737}$

$N_0$ = 6.78 g

So, 6.78 g of the radioactive substance needs to be ordered.

4 0

4 years ago

A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9

mamaluj [8]

Answer:

i) amount of MSW generated:

Laguna: 19200 Kg/day

Makati: 38000 Kg/day

ii) number of trucks to collect twice weekly:

Laguna: 3 trucks

Makati: 5 trucks

iii) volume of MSW in tons that enter landfill/week:

Laguna: 147.84 ton/week

Makati: 292 ton/week

Explanation:

i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day

⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week

Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day

⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week

ii) truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg

⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg

⇒ loads/day = 2 * 3000 kg = 6000 Kg

⇒ operate/week = 5 * 6000 Kg = 30000 Kg

∴ Laguna: number of trucks needed/week= 134400 / 30000 = 4.48 ≅ 5 trucks

⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks

∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks

⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks

iii) enter landfill/week:

Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week

Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week

4 0

3 years ago

The weak base ionization constant (Kb) for CN is equal to: