All categories

2 years ago

What part of the microscope is the letter k?

Chemistry

1 answer:

snow_tiger [21]2 years ago

6 0

Answer:

if you are asking k then the round one is condenser

if not then its a stage clip

You might be interested in

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

2 years ago

You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th

ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially, only 4 moles of oxygen gas were present in the flask.

$p_{O_2}=Tp_1\times X_{O_2}$ ( $X_{O_2}=\frac{4}{4}$ ) ( according to Dalton's law of partial pressure)

$p_{O_2}=Tp_1\times 1=Tp_1$ ....(1)

$Tp_1$ = Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

$X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}$

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

$p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

$Tp_2$ = Total pressure of the mixture.

from (1)

$Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

On rearranging, we get:

$Tp_2=2\times Tp_1$

The new total pressure will be twice of initial total pressure.

7 0

2 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion.

Alina [70]

Answer: The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_4^{2-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2$

Dividing 2 from 3, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2$

3 0

3 years ago

Which list of elements from Group 2 on the Periodic Table is arranged in order of increasing atomic radius?

IgorC [24]

The correct answer is option 1. Be, Mg, and Ca is the correct order arranged in increasing atomic radius. This is predicted based on the periodic table. The atomic sizes increases as one moves downwards in the periodic table.

4 0

2 years ago

A box is 1 m high, 2.5 m long, and 1.5 m wide, its volume is 5 mº. true or false

Maurinko [17]

Answer:

false

Explanation:

Because volume is in m3

5 0

2 years ago

Read 2 more answers