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3 years ago

Which pair of elements would most likely combine to form a salt? I and F Cs and I Na and C N and F

Chemistry

1 answer:

Savatey [412]3 years ago

7 0

Answer:

Cs and I

Explanation:

Salts are formed when an ionic bond is formed between two elements in the compound. Let us recall that the kind of bond formed between any two elements depends on the magnitude of electronegativity difference between the two elements.

Among the options listed, the highest degree of electronegativity difference occurs for the bond between Cs and I. This implies that this bond is ionic and the combination of the two elements will lead to salt formation.

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How i the ma number of an atom calculated? (1 point)

Bess [88]

Proton plus neutron is the correct answer. Protons and neutrons have a mass of 1 and electrons have a mass of 0. So in order to find the mass of an atom you need to add the number of protons and the number of neutrons.

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1 year ago

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OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

6 0

3 years ago

Read 2 more answers

A student measures the mass of a 10m^3 block of gold to be 500 kg what is the density of the gold

Anuta_ua [19.1K]

Answer:

d = 50 kg/m³

Explanation:

density = $\frac{mass}{volume}$

mass = 500 kg

volume = 10m³

$\frac{500}{10}$ = 50

d = 50 kg/m³

8 0

3 years ago

Magnesium ribbon reacts vigorously with hydrochloric acid. The test tube gets hot and bubbles are produced. When a metal reacts

oee [108]

The answer is C, hydrogen gas. This is because in single replacement reactions, the single element (here Magnesium) replaces whichever element in the compound it corresponds to. Because Mg loses electrons since it’s a metal, it will replace the element which also loses electrons, which is Hydrogen here. So when they switch places, MgCl2 and H2 are made— and H2 is the hydrogen gas.

6 0

3 years ago

Read 2 more answers

How long, in seconds, would it take for the concentration of A to decrease from 0.860 M to 0.310 M?

kogti [31]

Answer:

2.038 seconds.

Explanation:

So, in the question above we are given the following parameters in order to solve this question. We are given a rate constant of 0.500 s^-, initial concentration= 0.860 M and final concentration= 0.310 M,the time,t =??.

Assuming that the equation for the first order of reaction is given below,that is;

A ---------------------------------> products.

Recall the formula below;

B= B° e^-kt.

Therefore, e^-kt = B/B°.

-kt = ln B/B°.

kt= ln B°/B.

Where B° and B are the amount of the initial concentration and the amount of the concentration remaining, k is the rate constant and t = time taken for the concentration to decrease.

So, we have; time taken,t = ln( 0.860/.310)/0.500.

==> ln 2.77/0.500.

==> time taken,t =2.038 seconds.

6 0

3 years ago