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1 year ago

Wet chemical system components typically are a storage tank or tanks, piping, nozzles, and a(an)?

1 answer:

5 0

Wet chemical system components typically are a storage tank or tanks, piping, nozzles, and an actuating mechanism.

<h3>Which parts make up an actuation mechanism?</h3>

The typical components of an actuation mechanism include a motor, transmission, control units, feedback, signaling, interlocking, and shutdown units. An electrical, hydraulic, or pneumatic drive is used to move a valve, shutter, or gate, which acts as an actuator to regulate the flow of fluids or gases.

<h3>What is the operation of the operator-dependent actuating mechanism?</h3>

When an automatic clutch is engaged, the control system 2 replaces the target-value signals coming over feeders 1 with the operator force signal if the operator uses the operator-dependent actuating mechanism.

Learn more about automatic clutch here:-

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You might be interested in

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin pla

RSB [31]

<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 25.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g$

To calculate the heat released by the reaction, we use the equation:

$q=mc\Delta T$

where,

q = heat released

m = Total mass = [1.25 + 25] = 26.25 g

c = heat capacity of water = 4.18 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(21.9-25.8)^oC=-3.9^oC$

Putting values in above equation, we get:

$q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ammonium nitrate = 1.25 g

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

$\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -0.428 kJ

n = number of moles = 0.0156 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol$

Hence, the enthalpy change of the reaction is -27. kJ/mol

4 0

3 years ago

Consider the following reaction: CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) The initial rate of the reaction is measured at several di

Ira Lisetskai [31]

Answer : The correct rate law for the reaction is,

$\text{Rate}=k[CHCl_3][Cl_2]^{1/2}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)$

Rate law expression for the reaction:

$\text{Rate}=k[CHCl_3]^a[Cl_2]^b$

where,

a = order with respect to $CHCl_3$

b = order with respect to $Cl_2$

Expression for rate law for first observation:

$0.0035=k(0.010)^a(0.010)^b$ ....(1)

Expression for rate law for second observation:

$0.0069=k(0.020)^a(0.010)^b$ ....(2)

Expression for rate law for third observation:

$0.0098=k(0.020)^a(0.020)^b$ ....(3)

Expression for rate law for fourth observation:

$0.027=k(0.040)^a(0.040)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1$

Dividing 2 from 3, we get:

$\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}$

Calculation used :

$1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}$

Thus, the rate law becomes:

$\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}$

7 0

3 years ago

What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP

Dvinal [7]

Vol.250 before its to much pressure

7 0

3 years ago

Read 2 more answers

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

3 years ago

PLZ ANSWER QUICKLY! The electronegativity values of carbon, hydrogen, and nitrogen are compared in the table. Comparison of Elec

andriy [413]

Answer:

c. CH4 < NH3 because the NH bond is more polar than the CH bond.

Explanation:

Actually, the electronegativity difference between carbon and hydrogen is just about 0.4. This meager difference in electronegativity corresponds to a nonpolar bond between the two atoms.

However, the electronegativity difference between nitrogen and hydrogen is about 0.9. This larger electronegativity difference corresponds to the existence of a polar covalent bond between the two atoms.

Hence the N-H bond is significantly polar unlike the C-H bond. This implies that CH4 molecules are only held together by weak dispersion forces while NH3 molecules are held together by stronger dipole-dipole interactions and hydrogen bonds.

8 0

2 years ago