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Georgia [21]
3 years ago
5

Chemical equation for sodium bromide​

Chemistry
1 answer:
Dafna1 [17]3 years ago
6 0
NaBr is the answer to what’s the chemical equation for sodium bromide
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The major elements of organic molecules is carbon identify three other carbon is normally bonded to
arsen [322]
Carbon normally bonded to Hydrogen, Nitrogen, Oxygen
4 0
3 years ago
Molarity of a salt water solution of "0.47" moles of NaCl dissolved in a volume of 0.25L
asambeis [7]

Answer:

1.88 M

Explanation:

The following data were obtained from the question:

Mole of NaCl = 0.47 mole

Volume of solution = 0.25L

Molarity =?

Molarity is defined as the mole of solute per unit litre of the solution. It can represented mathematically as:

Molarity = mole /Volume

Using the above formula, the molarity of the salt water solution can be obtained as follow:

Molarity = 0.47/0.25

Molarity = 1.88 M

5 0
2 years ago
Simplify the numercial expression 13-2x5
Tema [17]

Answer:

3

Explanation:

2x5 is 10 then 13-10 is 3

4 0
3 years ago
Read 2 more answers
The compound cisplatin has been extensively studied as an antitumor agent. It is synthesized by the following, unbalanced reacti
zimovet [89]
The balanced chemical equation would be as follows:

<span>K2PtCl4(aq) + 2NH3(aq) --> Pt(NH3)2Cl2(s) + 2KCl(aq)

We are given the amount of </span>K2PtCl4 to be used in the reaction. This will be the starting point for our calculations. We do as follows:

65 g K2PtCl4 ( 1 mol / 415.09 g ) ( 1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl ) ( 300.051 g / 1 mol ) = 46.99 g Pt(NH3)2Cl produced 
6 0
3 years ago
Read 2 more answers
What is the density of a liquid if 12.5 ml of the liquid has a density of 9.80 g?
blagie [28]

Answer:

The answer is

<h2>0.784 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>density =  \frac{mass}{volume}</h3>

From the question

mass of liquid = 9.8 g

volume = 12.5 mL

The density is

density =  \frac{9.8}{12.5}

We have the final answer as

<h3>0.784 g/mL</h3>

Hope this helps you

3 0
3 years ago
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