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2 years ago

What is the density of mass=108g volume=5.64cm3

Chemistry

1 answer:

nordsb [41]2 years ago

6 0

Answer:

19.15kg/m³

Explanation:

density=mass/volume

d=m/v

d=108/5.64

19.15kg/m³

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It is found that a gas undergoes a first-order decomposition reaction. If the rate constant for this reaction is 8.1 x 10-2 /min

kap26 [50]

Answer:

28.43 min

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $8.1\times 10^{-2}$ min⁻¹

Initial concentration $[A_0]$ = 0.1 M

Final concentration $[A_t]$ = $1.0\times 10^{-2}$ M

Time = ?

Applying in the above equation, we get that:-

$1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}$

$0.1e^{-8.1\times \:10^{-2}t}=10^{-2}$

$e^{-8.1\times \:10^{-2}t}=\frac{1}{10}$

$\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)$

$t=28.43\ min$

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3 years ago

What is Industrial chemistry?

mel-nik [20]

Answer:

study of chemical used in industries

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2 years ago

How many molecules of SrCrO4 are in a sample of SrCrO4 0.556 moles?

Amiraneli [1.4K]

Answer:

$3.35*10^{23}\ SrCrO_4\ molecules$

Explanation:

$We\ are\ given\ that,\\No.\ of\ moles\ of\ SrCrO_4=0.556\\Hence,\\As\ we\ know\ that,\\No.\ of\ particles=Avagadro's\ Constant*No.\ of\ moles\\We\ already\ know\ that\ Avagadro's\ Constant=6.022*10^{23}\\Here,\\No.\ of\ SrCrO_4\ molecules= 6.022*10^{23}*0.556\\Hence,\\No.\ of\ SrCrO_4\ molecules=3.348*10^{23} molecules\ \approx 3.35*10^{23}\ SrCrO_4\ molecules$

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2 years ago

Which Kelvin temperatures represent, respectively, the normal freezing point and the normal boiling point of water?

zhenek [66]

Answer:

4. 273 K and 373K.

Explanation:

Conversion from Celsius to Kelvin ..

Kelvin = Celsius + 273

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3 years ago

How many grams are in 0.572 moles of glucose, C6H12O6

LUCKY_DIMON [66]

Answer:

103.00

Explanation:

1 mole of Glucose

6 C = 6 * 12 = 72 grams

12H = 12 * 1 = 12 grams

6O = 6 * 16 =96 grams

Total = 180 grams

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1 mol of glucose = 180 grams

0.572 mols of glucose = x

x = 0.572 * 180

x = 103.00 grams

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2 years ago