What goes along with the "drawing conclusions" step in the scientific method?

Is the following nuclear equation balanced? Yes or No.

Maru [420]

Answer:

No

Reason:

During a beta decay, one neutron convert into one proton and one electron. The proton remained in nucleus while electron move away known as beta radiation. So atomic mass will be remained same but atomic no will increase by one. So correct ans will be with 218, 86 Po and beta radiation.

3 0

2 years ago

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In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,

Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - -

At t =equilibrium (x-s) s 2s

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$K_{sp}=s\times {2s}^2$

$K_{sp}=4s^3$

Given s = 7.4×10⁻³ M

So, Ksp is:

$K_{sp}=4\times (7.4\times 10^{-3})^3$

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2$

Solving for s', we get

s' = 4.0×10⁻⁵ M

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

3 0

3 years ago

If you poor two ionic colloids into the same test tube, what would the reaction most likely be?

-BARSIC- [3]

The answer is B implosion.

4 0

3 years ago

State evidence from the bright-line spectra that indicates element A is not present in

Schach [20]

After studying the graph,the evidence from the bright line spectra that indicates element A is not present in the mixture is 11 the lines of A are not present in the mixture 700 nm is missing.

4 0

3 years ago

29.2 kJ of heat is required to convert an 50.0 g sample of

hram777 [196]

Answer:

The heat of vaporization of acetaldehyde is 25.72 kJ/mol.

Explanation:

Mass of acetaldehyde = 50.0 g

Molar mass of acetaldehyde = 44.05 g/mol

Moles of acetaldehyde = $\frac{50.0 g}{44.05 g/mol}=1.1351 mol$

Amount of energy required to vaporize 50.0 grams of acetaldehyde = Q

Q = 29.2 kJ

The heat of vaporization of acetaldehyde :

$\Delta H_{vap}=\frac{Q}{n}=\frac{29.2 kJ}{1.1351 mol}=25.72 kJ/mol$

8 0

3 years ago