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2 years ago

The energy transformations are similar because they both involve transformations that

Chemistry

1 answer:

blondinia [14]2 years ago

3 0

Sorry but I don’t understand the question. Sorry I’m not any help

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Elements that do not have full outer electron shells will donate, share, or take from other atoms. Choose the items that have th

Gnesinka [82]

Answer:

Ok...

Answer: YTH/JYG

Explanation:

AI+ZN+H=YTH

7 0

3 years ago

Plz answer question! Due Today!!! Question is in picture! Will make BRAINLIEST for first correct answer!!! No links plz!

iris [78.8K]

Answer:

I'd Go. B. weigh everything, let the reaction happen, then weigh everything again.

3 0

3 years ago

Read 2 more answers

Classify the following reaction.

Klio2033 [76]

Answer:

Explanation:

combustion reacts with oxygen

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3 years ago

The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc

jek_recluse [69]

Answer: $K_{sp}=1.25\times 10^{-14}$

$pK_{sp}=13.90$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of magnesium phosphate is given as:

$Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}$

When the solubility of $Mg_3(PO_4)_2$ is S moles/liter, then the solubility of $Mg^{2+}$ will be 3S moles\liter and solubility of $PO_4^{3-}$ will be 2S moles/liter.

Thus S = 0.173 g/L or $\frac{0.173g/L}{262.8g/mol}=0.00065mol/L$

$K_{sp}=(3S)^3\times (2S)^2$

$K_{sp}=108S^5$

$K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}$

$pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90$

5 0

3 years ago

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago