Which of the following are saturated hydrocarbons?

Which statement about the balanced equations for nuclear and chemical changes is correct? (1 point)

iris [78.8K]

The true statement about the balanced equations for nuclear and chemical changes is; both are balanced according to the total mass before and after the change.

A basic law in science is called the law of conservation of mass. Its general statement is that mass can neither be created nor destroyed.

Both in chemical and nuclear changes, mass is involved and in both cases, the law of conservation of mass strictly applies.

This means that for both chemical and nuclear changes; total mass before reaction must be equal to total mass after reaction.

Hence, both reactions are balanced according to the total mass before and after the change.

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3 0

3 years ago

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Put the following steps of specimen preparation and staining in order: I. Application of staining dyes II. Heat fixation III. Sm

MrRissso [65]

Answer:

Explanation:

II

3 0

3 years ago

Calculate the molarity of hydroxide ion in an aqueous solution that has a poh of 5.00

Simora [160]

Answer:

1.00 x 10^-5 M

Explanation:

pOH=-log[OH^-]

so

[OH^-]=10^(-pOH)

=10^-5

=1.00 x 10^-5 M

4 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

Draw the curved arrow mechanism for the reaction between (2S,3S)-3,5-dimethylhexan-2-ol and PCl3. Note the specific instructions

Vikki [24]

Answer:

(2R,3S)-2-chloro-3,5-dimethylhexane

Explanation:

As first step we have the attack of the OH group to the P atom in the PCl3 and one of the Cl atoms would leave. Then we will have a rearrangement to produce a double bond with the oyxgen on the OH. Finally the Cl produced will attack the carbon in a Sn2 substitution reaction to produce the halide with an opposite configuration.

4 0

3 years ago