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adelina 88 [10]
2 years ago
13

The ratification of the covenant in Deuteronomy 27-30 includes ________. A. Covenant blessings and curses B. Messianic promises

C. Instructions for priests
Chemistry
1 answer:
mamaluj [8]2 years ago
4 0

Answer:Covenant blessings and curses

Explanation:

The scripture quoted in Deuteronomy 27-30 was Moses's ratification of the covenant between God and the people of Israel. He itemized the terms of the covenant and the blessings that follow obedience to the Lord and the curses that follow disobedience to the Lord. It is a statement of blessings and curses.

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The average concentration of bromde ion in seawater is 65 mg of bromide ion per kg of seawater. what is the molarity of the brom
allsm [11]
Usually concentrations are expressed as molarity, or moles of solute per liter solution. First, convert the mass of bromide ion to moles. The molar mass of bromine is 79.904 g/mol.

Moles of bromine = 65 mg * 1 g/1000 mg * 1 mol/79.904 g = 8.135×10⁻⁴ moles

Next, convert the mass of seawater to volume using the density.

Volume of seawater =  1 kg * 1 m³/ 1,025 kg * 1000 L/1 m³ = 0.976 L

Thus,
Molarity = 8.135×10⁻⁴ moles/0.976 L = 8.335×10⁻⁴ M
5 0
2 years ago
A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m
ioda
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = \frac{PV}{nT} = constant
\frac{P1 V1}{n T} = \frac{P2 V2}{n T}
==> P1 * V1 = P2 * V2
V2 = \frac{P1 V1}{P2} = \frac{445 0.016}{1225}
V2 = 0.00581 m3 = 5.81 L


7 0
2 years ago
x g of metal (equivalent wt=12) was completely dissolved in 100 cc of N/2 HCl. The volume was then made upto 500 cc.25 cc of thi
ad-work [718]

Answer:

zxzx

Explanation:

8 0
2 years ago
Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of
dybincka [34]

<u>Answer:</u> The electronic configuration of the elements are written below.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

  • <u>Option a:</u>  Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is 1s^22s^22p^2

  • <u>Option b:</u>  Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is 1s^22s^22p^63s^23p^3

  • <u>Option c:</u>  Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is 1s^22s^22p^63s^23p^64s^23d^3

  • <u>Option d:</u>  Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3

  • <u>Option e:</u>  Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6

Hence, the electronic configuration of the elements are written above.

4 0
2 years ago
The combustion of one mole of liquid ethanol, CH3CH2OH, produces 1367 kJ of heat. Calculate how much heat is produced when 235.0
deff fn [24]

Answer:- 6984 kJ of heat is produced.

Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.

Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:

235.0g(\frac{1mole}{46g})(\frac{1367 kJ}{1mole})

= 6984 kJ

So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.

6 0
2 years ago
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