All categories

4 years ago

Can someone explain to me what the element Americium is?

1 answer:

7 0

It's a radioactive chemical element. It's symbol is AM and the atomic number is 95. It is a transuranic member of the actinide series, in the periodic table located under the lanthanide element europium, and thus by analogy was named after the Americas.

You might be interested in

Can beans grow inside water

fredd [130]

No, They need something to hold on to, such as dirt

8 0

4 years ago

Read 2 more answers

Who want to be the brainliest

GrogVix [38]

Answer:

You cross multiply the valencies as shown

Explanation:

1. Al(OH)3

2. The valency of Calcium is 2- not 1- so,

CaO

3 0

3 years ago

What is the mass of 5.80x10^23 atoms of silver

Natasha_Volkova [10]

Answer:

follow me and mark me brainliest

6 0

3 years ago

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

4 years ago

Phosphorus combines with oxygen to form two oxides. Find the empirical formula for each oxide of phosphorus if the percentage co

Bond [772]

We are given with the percentage composition of oxygen in an oxide from phosphorus and oxygen equal to 43.6 percent. that is the phosphorus is equal to 56.4 percent. we divide each with the molar mass and divide each with the smaller quantity. P then is equal to 1 and O is equal to 1.5. In this case, the empirical formula is P2O3.

4 0

3 years ago