The density of sea water is 1.029 kilograms per liter. What is the density of

Review the noble gases in the periodic table. As molar mass increases, the number of electrons increases, resulting in

docker41 [41]

greater dispersion forces is in fact correct

5 0

3 years ago

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340 438 g how do I determine the number of significant figures in the measurement?

il63 [147K]

In determining significant figure, always remember the rules:
Zeros place in the beginning and end of a non-zero numbers are not significant
Zeros place in the middle of a non-zero numbers are considered significant.

Therefore, 340 438 grams has 6 significant figures.

3 0

3 years ago

Calculate Horxn for the following reaction:H3AsO4(aq) + 4 H2(g) --> AsH3(g) + 4 H2O(l)(Hof [AsH3(g)] = 66.4 kJ/mol; Hof [H

Juli2301 [7.4K]

Answer:

The Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = (-172.2) \, kJ$

Explanation:

Given- Standard Heat of Formation:

$\Delta H_{f}^{\circ } [H_{3}AsO_{4}(aq)]$ = -904.6 kJ/mol

$\Delta H_{f}^{\circ } [H_{2}(g)]$ = 0 kJ/mol,

$\Delta H_{f}^{\circ } [AsH_{3}(g)]$ = +66.4 kJ/mol

$\Delta H_{f}^{\circ } [H_{2}O(l)]$ = -285.8 kJ/mol

Given chemical reaction: H₃AsO₄(aq) + 4H₂(g) → AsH₃(g) + 4H₂O(l)

The standard enthalpy of reaction: $\Delta H_{r}^{\circ }$ = ?

To calculate the Standard enthalpy of reaction ( $\Delta H_{r}^{\circ }$ ), we use the equation:

$\Delta H_{r}^{\circ } = \sum \nu .\Delta H_{f}^{\circ }(products)-\sum \nu .\Delta H_{f}^{\circ }(reactants)$

$\Delta H_{r}^{\circ } = [1 \times \Delta H_{f}^{\circ } [AsH_{3} (g)] + 4 \times \Delta H_{f}^{\circ } [H_{2}O(l)]] - [1 \times \Delta H_{f}^{\circ } [H_{3}AsO_{4}(aq)] + 4 \times \Delta H_{f}^{\circ } [H_{2}(g)]$

$\Rightarrow \Delta H_{r}^{\circ } = [1 \times (+66.4\,kJ/mol) + 4 \times (-285.8\,kJ/mol) ] - [1 \times (-904.6\,kJ/mol) + 4 \times (0\,kJ/mol)]$

$\Rightarrow \Delta H_{r}^{\circ } = [-1076.8\, kJ] - [-904.6\,kJ]$

$\Rightarrow \Delta H_{r}^{\circ } = (-172.2 \, kJ)$

Therefore, the Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = (-172.2) \, kJ$

5 0

3 years ago

What will the volume of Neon gas be if 1.2 moles is being held at 3.4 atm

andrezito [222]

i think the answer is b bc it makes sense

7 0

3 years ago

What is the decay constant of 23892u whose half-life is 4.5×109year?

Mademuasel [1]

Answer is: decay constant for uranium-238 is 1,54·10⁻¹⁰ 1/y.
The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration.
t1/2 = 4.5·10⁹ y.
λ = 0,693 ÷ t1/2.
λ= 0,693 ÷ 4.5·10⁹ y.
λ = 1,54·10⁻¹⁰ 1/y.
λ - decay constant.
Decay constant is proportionality between the size of a number of radioactive atoms (in this example uranium) and the rate at which the number of radioactive uranium atoms decreases because of radioactive decay.

6 0

3 years ago