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Phantasy [73]
3 years ago
15

Predict the products of hydrobromic acid, H B r , reacting with magnesium hydroxide, M g ( O H ) 2 . Select one or more:

Chemistry
1 answer:
Aliun [14]3 years ago
6 0

Answer:

Magnesium bromide, MgBr2, and water, H2O.

Explanation:

Hello!

In this case, since HBr is an acid and Mg(OH)2 is a base, an acid-base reaction is undergone, by which a salt and water are produced as neutralization products:

HBr+Mg(OH)_2\rightarrow MgBr_2+H_2O

However, it need to be balanced since two bromine atoms are produced, therefore we write:

2HBr+Mg(OH)_2\rightarrow MgBr_2+2H_2O

Thus, the products are magnesium bromide, MgBr2, and water, H2O.

Best regards!

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It takes 500 kJ to remove one mole of electrons from the atoms at the surface of a solid metal. How much energy does it take to
prisoha [69]

Answer:

8.31 × 10⁻²² kJ

Explanation:

Step 1: Given data

Energy required to remove one mole of electrons from the atoms at the surface of a solid metal: 500 kJ/mol e⁻

Step 2: Calculate how much energy does it take to remove a single electron from an atom at the surface of this solid metal

We will use Avogadro's number: there are 6.02 × 10²³ electrons in 1 mole of electrons.

500 kJ/mol e⁻ × 1 mol e⁻/6.02 × 10²³ e⁻ = 8.31 × 10⁻²² kJ/e⁻

7 0
3 years ago
If an object has a mass of 243.8 g and occupies a volume of 0.125L, what will be the density of this object in gm/cm3?
Nonamiya [84]

Answer:

it might reach to 4g per L

3 0
3 years ago
The movement of fluids up and down in a cycle because of convection is_____
kirill [66]
Called a(n) convection current
5 0
3 years ago
The heat of combustion per mole for acetylene, C2H2(g), is -1299.5 kJ/mol. Assuming that the combustion products are CO2(g) and
guapka [62]

Answer: The heat of combustion per mole for acetylene is 227.7 kJ/mol.

Explanation:

The combustion equation of acetylene is as follows.

C_{2}H_{2} + \frac{5}{2}O_{2} \rightarrow 2CO_{2} + H_{2}O

Formula to calculate enthalpy of formation for a reaction is as follows.

\Delta H^{o}_{rxn} = \sum \Delta H_{products} - \sum \Delta H_{reactants}\\\Delta H^{o}_{rxn} = [2\Delta H^{o}_{f}(CO_{2}) + \Delta H^{o}_{f} (H_{2}O)] - [\Delta H^{o}_{f}(C_{2}H_{2}) + \frac{5}{2} \Delta H^{o}_{f} O_{2}]\\-1299.5 = 2(-393.5) + (-285.8) - \Delta H^{o}_{f} (C_{2}H_{2})\\\Delta H^{o}_{f} (C_{2}H_{2}) = 227.7 kJ/mol

Thus, we can conclude that heat of combustion per mole for acetylene is 227.7 kJ/mol.

3 0
3 years ago
The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.
exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

8 0
3 years ago
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