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2 years ago

Which of the following is an electrolyte? a. BaSO4 b. CaS c. AlPO4 d. SrCrO4

Chemistry

2 answers:

lora16 [44]2 years ago

8 0

Answer:

Explanation:

faltersainse [42]2 years ago

5 0

A that the awnser dnjdjdjdj

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The pH scale is called the logarithmic scale what does this mean?

Ratling [72]

A logarithmic scale is a nonlinear scale used when there is a large range of quantities

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3 years ago

Where y and x are concentrations of the peptide in the extract and raffinate, respectively in g/L. It is desired to extract at l

zubka84 [21]

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2 years ago

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:-

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

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

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

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

$t=28.43\ min$

3 0

3 years ago

What is the answer ?

Fantom [35]

The answer is C Does it propaget

4 0

2 years ago

Read 2 more answers

Find the volume of a gas STP if it’s volume is 80.0 mL at 109 kpa and-12.5c

photoshop1234 [79]

Answer:

Approximately $8.38 \times 10^8\; \rm mL$ , which is the same as $8.38 \times 10^5 \; \rm L$ . Assumption: the behavior of this gas is ideal.

Explanation:

Initial state of the gas:

$T_\text{initial} = -12.5\; \rm ^\circ C = (-12.5 + 273.15)\; \rm K = 260.65\; \rm K$ .
$P_\text{initial} = \; \rm 10^9\; \rm kPa = 10^{12}\; \rm Pa$ .

STP state:

$T_\text{STP} = 0\; \rm ^\circ C = 273.15\; \rm K$ .
$P_\text{STP} = 10^5\; \rm Pa$ .

The initial state of this gas can be changed to STP state in two steps:

First, reduce the pressure from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ .
Second, increase the temperature from $260.65\; \rm K$ to $273.15\; \rm K$ .

Assume that the gas acts like an ideal gas at all time. Also, assume that the number of gas particles did not change.

Assume that temperature stays the same when the pressure changes from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ . By Boyle's Law, volume is inversely proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{intermediate} &= V_\text{initial} \cdot \displaystyle \frac{P_{\text{initial}}}{P_{\text{STP}}} \\ &= 80.0\; \rm mL \times \frac{10^{12}\; \rm Pa}{10^5\; \rm Pa} = 8.00 \times 10^8\; \rm mL\end{aligned}$ .

After that, assume that pressure stays the same when the temperature changes from $\rm 260.65\; \rm K$ to $\rm 273.15\; \rm K$ . By Charles's Law, volume is proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{STP} &= V_\text{intermediate} \cdot \displaystyle \frac{T_{\text{STP}}}{T_{\text{initial}}} \\ &= 8.00 \times 10^8\; \rm mL \times \frac{273.15\; \rm K}{260.65\; \rm K} \\ &\approx 8.38\times 10^8\; \rm mL = 8.38 \times 10^5\; \rm L\end{aligned}$ .

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3 years ago