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3 years ago

What are the units of molar mass? A. L/g B. mol/g C. g/L D. g/mol SUBST

Chemistry

2 answers:

rosijanka [135]3 years ago

7 0

Answer:

D - g/mol

step by step method

den301095 [7]3 years ago

5 0

D
I think its the best answer

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An experimenter found that for the hydrolysis of lactose, ΔH° = +0.44 kJ mol–1 and ΔS° = +0.031 kJ mol–1 K–1. ΔG°for this reacti

Ksivusya [100]

Answer:

ΔG° = -8.8 kJ/mol

Explanation:

The standard Gibbs free energy of reaction (ΔG°) can be calculated using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

ΔH°: standard enthalpy of reaction

T: absolute temperature

ΔS°: standard entropy of reaction

At 298 K (the temperature that is usually used), ΔG° for the hydrolysis of lactose is:

ΔG° = ΔH° - T.ΔS°

ΔG° = 0.44 kJ/mol - 298 K × 0.031 kJ/mol.K

ΔG° = -8.8 kJ/mol

7 0

3 years ago

Give three examples of convection. PLEASE I AM BEING TIMED!!!

SIZIF [17.4K]

Boiling water, steam from a cup of tea and ice melting

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3 years ago

BIG POINTS! NO SPAM PLZ! WILL AWARD BRAINLIEST!

pentagon [3]

Answer:

1.0 M is the concentration of hydrochloric acid.

Explanation:

$HCl (aq) + NaOH (aq) \rightarrow H_2O (l) + NaCl (aq)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=2.0 M\\V_2=25.0 mL$

Putting values in above equation, we get:

$1\times M_1\times 50.0 mL=1\times 2.0 M\times 25.0 mL$

$M_1=\frac{1\times 2.0 M\times 25 mL}{1\times 50.0 mL}=1.0M$

1.0 M is the concentration of hydrochloric acid.

6 0

3 years ago

$5,000 multiplyed by $150 multiplyed by $2500 multiplyed by 3%

ikadub [295]

Answer:

I am not sure but I will say 5625

Explanation:

6 0

3 years ago

Read 2 more answers

The isomerization of cyclopropane (CP) to propene (P) is a first order process. At 760 K, 15% of a sample of cyclopropane change

Advocard [28]

<u>Answer:</u> The half life of cyclopropane at 760 K is 28.99 minutes.

<u>Explanation:</u>

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken for decay process = 6.8 minutes

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 15) = 85 grams

Putting values in above equation, we get:

$k=\frac{2.303}{6.8min}\log\frac{100}{85}\\\\k=2.39\times 10^{-2}min^{-1}$

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = ?

k = rate constant of the reaction = $2.39\times 10^{-2}min^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{2.39\times 10^{-2}min^{-1}}=28.99min$

Hence, the half life of cyclopropane at 760 K is 28.99 minutes.

6 0

3 years ago