CHEMISTRY COLLEGE

Calculate the mass in grams for 0.251 moles of Na2CO3

Answers

Answer 1

Answer:

Answer:

Explanation:

the molar mass for Na2CO3 is 2*23+12+3*16=106 g/mole

106*0.251=26.606 grames

Related Questions

Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.1) Calculate q for the reaction. You must show your work.2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.3) Calculate ΔH for the reaction in kJ/mol. You must show your work.
What other reactions is taking place?
How many unpaid electrons are in 1s2 2s2 2p6 3s2 3p6
What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255?
Enter the oxidation number of one atom of each element in each reactant and product.CH4(g)+2O2(g)⟶CO2(g)+2H2O(g)C in CH4 :H in CH4 :O in O2 :C in CO2 :O in CO2 :H in H2O :O in H2O :Which atom is reduced?Which atom is oxidized?

What is the change in electrons for nitrogen in the following reaction?S + NO3 - -> SO2 + NO

A. Gain 2 electrons
B. Gain 3 electrons
C. Lose 2 electrons
D. Lose 3 electrons

Answers

Nitrogen changes from +5 in $NO_3$ - to +2 in NO. This means nitrogen has gained 3 electrons. Option B

To determine the change in electrons for nitrogen in the given reaction, we need to compare the oxidation state of nitrogen in the reactant ( $NO_3$ -) and the product (NO).

In the reactant, -, nitrogen is in the +5 oxidation state. This is because oxygen has an oxidation state of -2, and there are three oxygen atoms in $NO_3$ -. Therefore, nitrogen must have an oxidation state of +5 to balance the overall charge of $NO_3$ -.

In the product, NO, nitrogen is in the +2 oxidation state. This is because oxygen has an oxidation state of -2, and there is only one oxygen atom in NO. Therefore, nitrogen must have an oxidation state of +2 to balance the overall charge of NO.

By comparing the oxidation states of nitrogen in the reactant and the product, we can determine the change in electrons. The change in oxidation state corresponds to the change in the number of electrons gained or lost by the nitrogen atom.

In this case, nitrogen changes from +5 in $NO_3$ - to +2 in NO. This means nitrogen has gained 3 electrons.

Therefore, the correct answer is B) Gain 3 electrons.

The nitrogen atom undergoes a reduction because it gains electrons, reducing its oxidation state from +5 to +2 in the reaction.

Option B

for more such question on electrons visit:

brainly.com/question/26084288

#SPJ8

The ph of coffee is 5.6. The ph of grapefruit juice is 2.6. This means that the proton concentration in coffee is

Answers

pH scale is used to measure the acidity or alkalinity on a scale of 0-14 where 0-6.9 is the acidic region , 7.1-14 is the basic region and 7 is for the neutral substance. We can calculate the concentration of proton from pH.

$pH=-log[H^+]$

where $[H^+]$ is the concentration of proton

As per the question ,the pH of coffee is 5.6 and we need to find the concentration of proton so putting the values in the above equation, we get

$5.6=-log[H^+]$

$-5.6=log[H^+]$

$antilog(-5.6)= [H^+]$

$[H^+]= 2.511*10^-^6 M$

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base (B) solution. Show all your work.pKb for the weak base = 4.82.

Molar mass of the weak base = 82.0343g/mole.

Note: pKa = -logKa

pKb = -logKb

pH + pOH = 14

[H+ ] [OH- ] = 10^-14

Answers

Answer:

11.39

Explanation:

Given that:

$pK_(b)=4.82$

$K_(b)=10^(-4.82)=1.5136* 10^(-5)$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = (Mass\ taken)/(Molar\ mass)$

Thus,

$Moles= (1.805\ g)/(82.0343\ g/mol)$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)$

$Molarity=(0.022)/(0.055)$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_(b)=\frac {\left [ BH^(+) \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136* 10^(-5)=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or negativea) Pb^2+(aq) + 2Cl-(aq) ---> PbCl2(s)b) CaCO3(s) ---> CaO(s) + CO2 (g)c) 2NH3(g) ---> N2(g) + 3H2(g)d) P4(g) + 5O2(g) ---> P4O10(s)e) C4H8(g) + 6O2(g) ---> 4CO2(g) + 4H2O(g)f) I2(s) ---> I2(g)

Answers

Answer: a) $Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$ : negative

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$ : positive

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$ : positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s)$ : negative

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$ : positive.

f) $I_2(s)\rightarrow I_2(g)$ : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

a) $Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$

As ions are moving to solid form , randomness decreases and thus sign of $\Delta S$ is negative.

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s)$

As gas is changing to solid, randomness decreases and thus sign of $\Delta S$ is negative.

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

f) $I_2(s)\rightarrow I_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

What is the hydrogen concentration of a pH of 6.95, 7.25 and 8.1?

Answers

Explanation:

pH is use calculate the acidic strength of a substance ,

The value of pH is calculated as , negative log of the concentration of Hydrogen ions .

Therefore ,

pH = - log [ H ⁺]

rearranging the above equation , to find the hydrogen ion concentration . log [ H ⁺] = pH

log [ H ⁺] = - pH

[ H ⁺] = antilog ( - pH )

Hence , from the question , the hydrogen ions can be calculated by using the above equation ,

( 1 )

pH = 6.95

Since ,

[ H ⁺] = antilog ( - pH )

[ H ⁺] = antilog ( - 6.95 )

[ H ⁺] = 1.12 * 10 ⁻⁷

( 2 )

pH = 7.25

Since ,

[ H ⁺] = antilog ( - pH )

[ H ⁺] = antilog ( - 7.25 )

[ H ⁺] = 5.62 * 10 ⁻⁸

( 3 )

pH = 8.1

Since ,

[ H ⁺] = antilog ( - pH )

[ H ⁺] = antilog ( - 8.1 )

[ H ⁺] = 7.94 * 10 ⁻⁹

Liquid ethyl mercaptan, C2H6S, has a density of 0.84 g/mL. Assuming that the combustion of this compound produces only CO2 , H2O, and SO2 , what masses of each of these three products would be produced in the combustion of 3.15 mL of ethyl mercaptan

Answers

Answer:

Mass CO2 = 3.75 grams

Mass H2O = 2.30 grams

Mass SO2 = 2.73 grams

Explanation:

Step 1: Data given

Density of Liquid ethyl mercaptan, C2H6S = 0.84 g/mL

Volume of ethyl mercaptan = 3.15 mL

Step 2: The reaction

2C2H6S + 9O2 → 4CO2 + 6H2O + 2SO2

Step 3: Calculate mass of ethyl mercaptan

Mass = Volume * density

Mass ethyl mercaptan = 3.15 mL * 0.84 g/mL

Mass ethyl mercaptan = 2.646 grams

Step 4: Calculate moles ethyl mercaptan

Moles = mass / molar mass

Moles ethyl mercaptan = 2.646 grams / 62.13 g/mol

Moles ethyl mercaptan = 0.04259 moles

Step 5: Calculate moles of other products

For 2 moles ethyl mercaptan we need 9 moles O2 to produce 4 moles CO2, 6 moles H2O and 2 moles SO2

For 0.04259 moles we need 0.1917 moles O2 to produce:

2*0.04259 = 0.08518 moles CO2

3*0.04259 = 0.1278 moles H2O

1*0.04259 = 0.04259 moles SO2

Step 6: Calculate mass produced

Mass = moles * molar mass

Mass CO2 = 0.08518 moles * 44.01 g/mol

Mass CO2 = 3.75 grams

Mass H2O = 0.1278 moles * 18.02 g/mol

Mass H2O = 2.30 grams

Mass SO2 = 0.04259 moles * 64.07 g/mol

Mass SO2 = 2.73 grams

Other Questions

In Natural selection, individuals whose unique characteristics are well-suited for an environment tend to survive and produce more offspring. true or false?

Which is an example of plasmas in nature? solar cells plasma balls auroras clouds

g n the following three compounds(1,2,3) arrange their relative reactivity towards the reagent CH3Cl / AlCl3. Justify your order

. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.

Calculate the mass in grams for 0.251 moles of Na2CO3￼

Answers

Related Questions

Answers

Answers

Answers

Answers

Answers

Answers

Calculate the mass in grams for 0.251 moles of Na2CO3