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Blababa [14]
3 years ago
13

Does anyone know where i can find a practice cumulative chem test online?

Chemistry
1 answer:
Montano1993 [528]3 years ago
6 0
Exam board websites often have past papers from several previous years, if you know what exam board you are going to be using, try that, if not there may be content on the edexcel and AQA websites
You might be interested in
A water treatment plant applies chlorine for disinfection so that 10 mg/L chlorine is achieved immediately after mixing. The vol
Vlada [557]

<u>Answer:</u> The mass of chlorine needed by the plant per day is 9.4625\times 10^8mg

<u>Explanation:</u>

We are given:

Volume o water treated per day = 25,000,000 gallons

Converting this volume from gallons to liters, we use the conversion factor:

1 gallon = 3.785 L

So, \frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L

Amount of chlorine applied for disinfection = 10 mg/L

Applying unitary method:

For 1 L of water, the amount of chlorine applied is 10 mg

So, for 9.4625\times 10^7L of water, the amount of chlorine applied will be \frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg

Hence, the mass of chlorine needed by the plant per day is 9.4625\times 10^8mg

6 0
3 years ago
What is the oxidation number of chlorine
sergiy2304 [10]
The oxidation number of chlorine in the Cl-<span> ion is -1.</span>
3 0
3 years ago
A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m
ioda
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = \frac{PV}{nT} = constant
\frac{P1 V1}{n T} = \frac{P2 V2}{n T}
==> P1 * V1 = P2 * V2
V2 = \frac{P1 V1}{P2} = \frac{445 0.016}{1225}
V2 = 0.00581 m3 = 5.81 L


7 0
4 years ago
Calculate the boiling point of a solution of 500.0 g of ethylene glycol (c2h6o2) dissolved in 500.0 g of water. kf = 1.86°c/m a
Bingel [31]

Answer:The boiling point of the solution is 108° C.

Explanation:

Boiling point of pure water=T=100^oC

Boiling point of water after addition of 500 g of ethylene glycol=T_f

Mass of water = 500g = 0.5 kg (1000 g = 1 kg)

\Delta T_f=K_b\times \frac{\text{Mass of ethlyene glycol}}{\text{Molar mass of ethylene glycol}}

\Delta T_f=0.512^oC/m\times \frac{500 g}{62.07 g/mol\times 0.5 kg}

\Delta T_f=8.24 ^oC

\Delta T_f=T_f-T

8.24^oC=T_f-100^oC

T_f=108.24^oC

The boiling point of the solution is 108° C.

7 0
3 years ago
A sample of nitrogen gas is stored in a 489.6 mL flask at 108 kPa and 10.0°C. The gas is transferred to a 750.0 mL flask at 28.7
Wewaii [24]

Answer:

P2= 125.26 Kpa

Explanation:

V1= 489.6 ml=0.4896L

V2= 750 ml= 0.750L

V1= 180 KPa= 180000 Pa

P2= ?

T1= 10 = 10 + 273.15 = 283.15K

T2= 28.7+273.15= 301.85K

180000Pa* 0.4896L/ 283.15K * 301.85K/0.75L

P2= 12526.28553

P2= 125.26 KPa

5 0
3 years ago
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