How many grams of Cu would be needed to react with 2.0 mol of HNO3?

Most modern medications are given in doses of milligrams. Thyroid medications, however, are typically given in doses of microgra

leonid [27]

Answer:

0.125 mg

Explanation:

The correct answer would be 0.125 mg

According to the conversion factor, one milligram of a sample is equivalent to one thousand micrograms of the same sample.

milligram = $10^{-3}$

microgram = $10^{-6}$

Hence,

1 milligram = 1000 micrograms or 1 microgram = $10^{-3}$ milligram

Therefore, 125 micrograms will be:

125/1000 = 0.125 milligram

4 0

3 years ago

Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con

Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is $K_a\times K_b$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) $P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)$

$K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}$

b) $PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)$

$K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}$

For overall reaction on adding a and b we get c

c) $P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)$

$K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}$

$K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}$

The equilibrium constant for the overall reaction is $K_a\times K_b$

4 0

3 years ago

1) Given the balance equation below. Calculate how much Na3PO4 in grams you

juin [17]

Answer:

136.67 g of Na3PO4 is required to create 100 gram of NaOH.

Explanation:

The balanced equation:

$Na_{3}PO_{4} + 3 KOH \rightarrow 3 NaOH + K_{3}PO_{4}$

1 mole Na3PO4 = 164 g/mole (Molar mass)

1 mole NaOH = 40 g/mole (Molar mass)

Now,

1 mole of Na3PO4 produce = 3 mole of NaOH

164 g/mol of Na3PO4 produce = 3(40) g/mol of NaOH

or

120 g/mol of NaOH is produced from = 164 g/mol of Na3PO4

1 g/mol of NaOH is produced from =

$\frac{164}{120}$

100 grams of NaOH is produced from =

$\frac{164}{120}\times100$ gram of Na3PO4

calculate,

= 136.67 g

8 0

3 years ago

Which ions aren't shown in net ionic reactions, because they are present as both reactants and products of the reaction, so they

Anvisha [2.4K]

Explanation:

This is correct!

Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.

An example is;

Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)

The ions; Na+, NO3−(aq) would be cancelled out to give;

Cl−(aq) + Ag+(aq) → AgCl(s)

7 0

3 years ago

Methanol can be produced by the following reaction: CO(g) 2 H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas rel

belka [17]

Answer:

$r_{H_2}=-2r_{CH_3OH}$

Explanation:

Hello!

In this case, for the reaction:

$CO(g)+ 2 H_2(g) \rightarrow CH_3OH(g)$

In such a way, via the rate proportions, that is written considering the stoichiometric coefficients, we obtain:

$\frac{1}{-1} r_{CO}=\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}$

Whereas the reactants, CO and H2 have negative stoichiometric coefficients; therefore the rate of disappearance of hydrogen gas is related to the rate of appearance of methanol as shown below:

$\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}\\\\r_{H_2}=\frac{-2}{1} r_{CH_3OH}\\\\r_{H_2}=-2r_{CH_3OH}$

Which means that the rate of disappearance of hydrogen gas is negative and the rate of appearance of methanol is positive.

Regards!

7 0

3 years ago