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Ronch [10]
2 years ago
15

The compound NaCl (salt) has the same properties as sodium (Na) and chlorine (CI) because it is made of both of those elements.

True False​
Chemistry
1 answer:
grin007 [14]2 years ago
3 0

Answer:

False.

Explanation:

Hello.

In this case since sodium is a soft brilliant metal and chlorine is a green gas at normal conditions, they have totally different properties. Thus, since sodium chloride is also known as the table salt which is present as white crystals we can infer that the compound does not have the same properties, therefore it is false.

Regards.

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How many grams of sodium chloride are present in a 0.75 M solution with a volume of 500.0 milliliters?
saveliy_v [14]
The molarity of a solution is a type of expression of concentration equal to the number of moles solute per liter solution. In this problem, we are given the molarity equal to 0.75 M and a volume equal to 500 milliliters. <span>500 milliliters is equal to 0.5 liters. we multiply M and L to get the number of moles then multiply by the molar mass of NaCl. The answer is 21.92 grams.</span>
6 0
3 years ago
If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th
qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

K_c=4.74\times 10^4

Concentration at equilibrium:

[COBr_2]=1.58\times 10^{-6}M

[Co]=2.78\times 10^{-3}M

[Br_2]=2.51\times 10^{-5}M

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0
3 years ago
What is the basic unit of chemistry?
Luba_88 [7]
An atom hopefully this helps
7 0
2 years ago
Read 2 more answers
Samples of each of the following objects are removed from a 60º C water bath, in which they have been for enough time to reach t
Brut [27]

Answer:

concrete

2kj for second question

Explanation:

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4 0
3 years ago
What volume of water is required to prepare 0.1 M H3PO4 from 100 ml of 0.5 M solution?
ExtremeBDS [4]

Answer: A volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

Explanation:

Given: M_{1} = 0.1 M,    V_{1} = ?

M_{2} = 0.5 M,       V_{2} = 100 mL

Formula used to calculate the volume of water is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

7 0
3 years ago
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