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3 years ago

Decide what the best solution would be if you reach your lab table and find a broken beaker with liquid spilling out.

1 answer:

6 0

There are many safety precautions and rules you MUST follow during labs.
for this incident here is what you should do:
1)Notify your Instructor and partner
2) if the liquid is toxic (like not water or vinegar) then let your Instructor handle it properly, or follow your instructors orders (like if they say to put a towel over it or something like that)
3) Broken glassware, minus mercury thermometer, must be immediately cleaned up, do not use your bare hand, always wear gloves.
4)dispose of the broken glass properly and clean the liquid up (unless it is harmful, then let your instructor do it)

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In an isolated system, mass can neither be<br> created or destroyed.<br> ube<br> FALSE<br> TRUE

Marina86 [1]

Answer:

True

Explanation:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

Hope it helps

5 0

3 years ago

Classify the following alcohol as primary,

Nata [24]

The classification of the alcohols gives;

Compound 1 - Primary alcohol
Compound 2 - Tertiary alcohol
Compound 3 - Secondary alcohol
Compound 4 - Secondary alcohol

<h3>What are alcohols?</h3>

Organic compounds occurs in families. The family of compounds is called a homologous series. The homologous series always have a functional group. The functional group is the atom, group of atoms or bond that is responsible for the chemical reactivity of the members of a given homologous series.

Now we know that the alcohols are those organic compounds that contains the -OH group. The could be aliphatic or alicyclic compounds. We shall now proceed to name the kind of alcohols that each of the compounds shown are;

Compound 1 - Primary alcohol
Compound 2 - Tertiary alcohol
Compound 3 - Secondary alcohol
Compound 4 - Secondary alcohol

Learn ore about alcohols:brainly.com/question/4698220

#SPJ1

3 0

1 year 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

The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall

miv72 [106K]

Answer:

The value is $C_p = 42. 8 J/K\cdot mol$

Explanation:

From the question we are told that

$\gamma = \frac{C_p }{C_v}$

The initial volume of the fluorocarbon gas is $V_1 = V$

The final volume of the fluorocarbon gas is $V_2 = 2V$

The initial temperature of the fluorocarbon gas is $T_1 = 298.15 K$

The final temperature of the fluorocarbon gas is $T_2 = 248.44 K$

The initial pressure is $P_1 = 202.94\ kPa$

The final pressure is $P_2 = 81.840\ kPa$

Generally the equation for adiabatically reversible expansion is mathematically represented as

$T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }$

Here R is the ideal gas constant with the value

$R = 8.314\ J/K \cdot mol$

$248.44 = 298.15 * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }$

=> $C_v = 31.54 J/K\cdot mol$

Generally adiabatic reversible expansion can also be mathematically expressed as

$P_2 V_2^{\gamma} = P_1 V_1^{\gamma}$

=> $[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}$

=> $2^{\gamma} = 2.56$

=> $\gamma = 1.356$

$\gamma = \frac{C_p}{C_v} \equiv 1.356 = \frac{C_p}{31.54}$

=> $C_p = 42. 8 J/K\cdot mol$

3 0

3 years ago

If 3.50 moles of sodium chloride is added to your food, how many grams is added?

Ksenya-84 [330]

Answer:

204.8g

Explanation:

The number of moles of a substance is related to its mass and molecular mass as follows:

mole (n) = mass (m) ÷ molar mass (MM)

According to this question, 3.50 moles of sodium chloride (NaCl) is added to a food.

Molar mass of NaCl = 23 + 35.5

= 58.5g/mol

Using mole = mass/molar mass

Mass = molar mass × mole

Mass = 58.5g/mol × 3.5mol

Mass = 204.75

Mass = 204.8grams.

Therefore, 204.8grams of NaCl or common salt was added to the food.

7 0

3 years ago