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3 years ago

What is the molarity of the solution produced when 8.0g NaOH is dissolved in sufficient water to prepare 500.mL of solution?

Chemistry

1 answer:

Sindrei [870]3 years ago

8 0

Answer:

Molarity = 0.4M

Explanation:

Molar mass of NaOH (M)= 40

m= 8g, V= 500ml=0.5L

n= m/M=[8/40]= 0.2mol

Applying

n= CV

0.2= C×0.5

C= 0.4M

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Please help!! Balancing Nuclear Equations

sineoko [7]

The missing part of the equation is found to be 4/2He. Option A

<h3>What are nuclear equations?</h3>

The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.

Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.

Learn more about nuclear equations:brainly.com/question/19752321

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8 0

2 years ago

Find mass of 3 moles of water

EleoNora [17]

Answer:

54 g

Explanation:

1 mole of water = H2O

mass of 1 mole of H2O= mass of h2 + mass of o

= 2× mass of h +mass of o

= 2×1+16 =18 g

1 mole of water = 18g

3moles of water = 18×3g= 54g

6 0

3 years ago

How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?

Dmitriy789 [7]

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

6 0

3 years ago

Please help ASAP!! <br><br>I am not sure how to solve this! <br><br>

umka2103 [35]

Answer:

Potassium (K) has 19 protons.

Each neutral atom of Potassium has 19 electrons.

Explanation:

A) Potassium has 19 protons because the atomic number tells us how many protons are in an atom of the element. (The atomic number is the number above the element symbol. For example, the number above "K" is 19, which is the atomic number).

B) If an atom is neutral, this means that the atom has neutral energy. Protons give positive energy and electrons give negative energy. For the atom to be neutral, the atom must have balanced energy, therefore, making the number of electrons equal to the number of protons in a neutral atom.

7 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago