Is my current answer correct? if not, please correct me

When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net

Sonja [21]

Answer:

$Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)$

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:

$Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)$

Now, we can separate the nitrates in ions as they are aqueous to obtain:

$Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)$

And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:

$Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)$

Best regards!

6 0

3 years ago

what quantity of ethylen glycol must be added to 125 g of water to raise the boiling point of 1.0 degrees celsuis

labwork [276]

Answer:

We should add 15.15 grams of ethylen glycol.

Explanation:

Step 1: Data given

Mass of water = 125 grams

Temperature change = 1.0 °C

The boiling point elevation constant, Kbp, for water is 0.5121 °C/m

Step 2: Calculate mass needed

ΔT = i* Kb * m

⇒ with i = the van't Hoff factor = 1

⇒ with Kb = 0.5121 °c/ kg/mol

⇒ with m = molality = moles / mass

ΔT = i* Kb * m ⇒ 1°C = 0.5121 °C / kg/mol * (X/0.125kg)

X = 0.2441 moles

Step 3: Calculate mass of ethylen glycol

Mass ethylen glycol = moles * molar mass

Mass ethylen glycol = 0.2441 moles * 62.068 g/mol

Mass of ethylen glycol = 15.15 grams

We should add 15.15 grams of ethylen glycol.

4 0

3 years ago

Number 14 please help.. chemistry

Zolol [24]

Taking into account the definition of density, the mass of a substance with a volume of 150 cm³ and a density of 1.5 $\frac{g}{cm^{3} }$ is 225 grams.

<h3>What is density</h3>

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

<h3>Mass of the substance in this case</h3>

In this case, you know that:

Density= 1.5 $\frac{g}{cm^{3} }$
Volume= 150 cm³

Replacing in the definition of density:

$1.5 \frac{g}{cm^{3} } =\frac{mass}{150 cm^{3} }$

Solving:

mass= 1.5 $\frac{g}{cm^{3} }$ ×150 cm³

In summary, the mass of a substance with a volume of 150 cm³ and a density of 1.5 $\frac{g}{cm^{3} }$ is 225 grams.

Learn more about density:

brainly.com/question/952755

brainly.com/question/1462554

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7 0

2 years ago

Which graph best demonstrates the general relationship between mass and temperature, similar to the trend of thermal energy abso

bulgar [2K]

It is the first one.

6 0

3 years ago

Read 2 more answers

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

Anna [14]

<u>Answer:</u> The number of moles of ethanol after equilibrium is reached the second time is 11. moles.

<u>Explanation:</u>

We are given:

Initial moles of ethene = 34 moles

Initial moles of water vapor = 15 moles

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 34 15

At eqllm: 34-x 15-x x

We are given:

Equilibrium moles of ethene = 24 moles

Equilibrium moles of water vapor = 5 moles

Calculating for 'x'. we get:

$34-x=24\\\\x=10$

Volume of container = 100.0 L

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}$ .......(1)

$[CH_3CH_2OH]=\frac{10}{100}=0.1M$

$[CH_2=CH_2]=\frac{24}{100}=0.24M$

$[H_2O]=\frac{5}{100}=0.05M$

Putting values in expression 1, we get:

$K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3$

Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 24+11 5 10

At eqllm: 35-x 5-x 10+x

$[CH_3CH_2OH]=\frac{10+x}{100}$

$[CH_2=CH_2]=\frac{35-x}{100}$

$[H_2O]=\frac{5-x}{100}$

Putting values of in expression 1, we get:

$8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1$

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.

Moles of ethanol = $(10+x)=10+1.1=11.$

Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.

8 0

3 years ago

Is my current answer correct? if not, please correct me​

Is my current answer correct? if not, please correct me