Which of the following descriptions best describes a weak base?

You have 125 g of a certain seasoning and are told that it contains 62.0 g of salt. what is the percentage of salt by mass in th

r-ruslan [8.4K]

62.0g/125g= 0.496 x 100 = 49.6%

4 0

2 years ago

What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?

Ilya [14]

Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

4 0

1 year ago

Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi

hoa [83]

Answer:

i) 0.5071 (kg/s)

ii) -1407.1 kj/kg

iii) 204.05 Kw

iv) 5.881

v) 9.238

Explanation:

Given Data:

evaporation temperature ( T ) = 4°c = 277.15 K

Condensation Temperature ( T ) = 34°c = 307.15 K

n ( compressor efficiency ) = 0.76

refrigeration rate = 1200 kJ.s^-1

i) determine the circulation rate of the refrigerant

m = $\frac{Q}{H2 - H1}$ = $\frac{Q}{H2 - H4\\}$ ------- 1

Q = 1200 Kj.s^-1

H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )

H4 = entropy at step 4 = 142.4 ( kJ/ kg )

back to equation 1

m ( circulation rate of refrigerant ) = 0.5071 (kg/s)

ii) heat transfer rate in the condenser

Q = m ( H4 - H3 )

= 0.5071 ( 142.4 - 2911.27 )

= -1407.1 kj/kg

where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )

iii) power requirement

w = m * ΔH23

= 0.5071 (kg/s) * 402.37 (kj/kg) = 204.05 Kw

where: ΔH23 = $\frac{H'3 - H2 }{0.76}$ = $\frac{2814.7-2508.9}{0.76}$ = 402.37 (kj/kg)

iv) coefficient of performance of a cycle

W = Qc / w

= 1200 Kj.s^-1/ 204.05 kw

= 5.881

v) coefficient of performance of a Carnot refrigeration cycle

$w_{carnot} = \frac{T2}{T4 - T2}$

= 277.15 / ( 307.15 - 277.15 )

= 9.238

4 0

2 years ago

What is the empirical formula of a substance that contains 0.117 mol of carbon, 0.233 mol of hydrogen, and 0.117 mol of oxygen?

Rzqust [24]

Empirical formula of a compound gives the proportions of the elements in that compound but it does not define the actual arrangement and number of atoms.

Let the empirical formula of compound be $C_{x}H_{y}O_{z}$ .

The ratio of number of moles of C, H and O can be calculated as follows:

$C:H:O=0.117:0.233:0.117$

Simplifying the ratio,

$C:H:O=\frac{0.117}{0.117}:\frac{0.233}{0.117}:\frac{0.117}{0.117}=1:1.99:1=1:2:1$

Thus, the value of x, y and z will be 1, 2 and 1 respectively.

Therefore, the empirical formula will be $CH_{2}O$ .

6 0

1 year ago

Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was or

givi [52]

Answer: The initial amount of Uranium-232 present is 11.3 grams.

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

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

We are given:

$t_{1/2}=68.9yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{68.9}=0.0101yr^{-1}$

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 = $0.0101yr^{-1}$

t = time taken for decay process = 206.7 yrs

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

$0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}$

$[A_o]=11.3g$

Hence, the initial amount of Uranium-232 present is 11.3 grams.

4 0

2 years ago

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