Ask question

Ask question

All categories

kakasveta [241]

3 years ago

6

Explain why Thorium-229 is radioactive

1 answer:

Vladimir79 [104]3 years ago

7 0

Answer:

Thorium-229 is a very insoluble metal that emits radioactive alpha particles. It is generated by the decay of uranium-233 and has a half-life of 7.340 years.

Explanation:

braliyest please

You might be interested in

Q05 the "atomic number" of an atom is determined by the number of ________ it has.

Alexxx [7]

The atomic number of an atom is determined by the number of protons it has..

It is also the whole number shown on the periodic table

6 0

4 years ago

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

Ahat [919]

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to determine the oxidation state of the elements in the compound.

(a) $H_2SO_4$

Let the oxidation state of 'S' be, 'x'

$2(+1)+x+4(-2)=0\\\\x=+6$

Hence, the oxidation state of 'S' is, (+6)

(b) $Ca(OH)_2$

Let the oxidation state of 'Ca' be, 'x'

$x+2(-2+1)=0\\\\x=+2$

Hence, the oxidation state of 'Ca' is, (+2)

(c) $BrOH$

Let the oxidation state of 'Br' be, 'x'

$x+(-2)+1=0\\\\x=+1$

Hence, the oxidation state of 'Br' is, (+1)

(d) $ClNO_2$

Let the oxidation state of 'N' be, 'x'

$-1+x+2(-2)=0\\\\x=+5$

Hence, the oxidation state of 'N' is, (+5)

(e) $TiCl_4$

Let the oxidation state of 'Ti' be, 'x'

$x+4(-1)=0\\\\x=+4$

Hence, the oxidation state of 'Ti' is, (+4)

(f) $NaH$

Let the oxidation state of 'Na' be, 'x'

$x+(-1)=0\\\\x=+1$

Hence, the oxidation state of 'Na' is, (+1)

4 0

3 years ago

ohaa [14]

Answer:

species originate through allopatric speciation:

Have a good day! :)

6 0

4 years ago

Read 2 more answers

A gas has a volume of 490. mL at a temperature of -35.0 degrees C. What volume would the gas occupy at 42.0 degrees Celsius? Ple

miskamm [114]

Answer:

648.5 mL

Explanation:

Here we will assume that the pressure of the gas is constant, since it is not given or specified.

Therefore, we can use Charle's law, which states that:

"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically:

$\frac{V}{T}=const.$

where

V is the volume of the gas

T is its absolute temperature

The equation can be rewritten as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where in this problem we have:

$V_1=490 mL$ is the initial volume of the gas

$T_1=-35.0^{\circ} + 273 = 238 K$ is the initial temperature

$T_2=42.0^{\circ}+273=315 K$ is the final temperature

Solving for V2, we find the final volume of the gas:

$V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL$

8 0

4 years ago

Consider a solution that contains 0.274 M potassium chloride and 0.155 M magnesium chloride.

solong [7]

Answer:

Concentration of chloride ions = 0.584M

Explanation:

The step by step calculations is shown as attached below.

3 0

4 years ago