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3 years ago

9

Calculate the concentration of the rubidium hydroxide solution in g dm³. (relative atomic mass: Rb = 85; relative formula mass:

RbOH = 102)

1 answer:

Fofino [41]3 years ago

5 0

Answer:

assalam o aalaikum

Explanation:

o

90

180

270

360 x

-1

Given that 0 find the value of a and the value of b.

(2 marks)

.

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Round off the measurement to three significant figures 12.17º C

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12.2 C
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How many neurons does atom in the model shown below have<br> a.2<br> b.9<br> c.10 <br> d.16

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Read 2 more answers

Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

8 0

3 years ago

The standard enthalpy change for the following reaction is 873 kJ at 298 K.

Flura [38]

Answer: - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

$2KCl(s)\rightarrow 2K(s)+Cl_2(g)$ $\Delta H_1=873kJ$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s)$ $\Delta H_2=?$

According to the Hess’s law, if we divide the reaction by half then the $\Delta H$ will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value $\Delta H_2$ for the reaction will be:

$\Delta H_2=\frac{1}{2}\times (-873kJ)$

$\Delta H_2=-436.5kJ$

Hence, the value of $\Delta H_2$ for the reaction is -436.5 kJ.

8 0

3 years ago