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3 years ago

12

the density of water at different temperature is listed in the table above. Based on this information we can predict that the de

nsity of water at 50 might be and the density at 150 might be?

1 answer:

son4ous [18]3 years ago

5 0

look it nup on google should help

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Substances that cause the oxidation of other substances are

svp [43]

<h3>SUBSTANCE</h3>

==============================

$\large \sf\underline{Question:}$

Substances that cause the oxidation of other substances are;

==============================

$\large \sf\underline{Answer:}$

$\qquad \qquad \huge \bold{Option \: B}$

==============================

$\large \sf\underline{Explanation:}$

Substances that cause the oxidation of other substances are oxidizing agents.

==============================

4 0

2 years ago

inessss [21]

Answer:

LiO2

Explanation:

One Lithium per 2 Oxygen

3 0

3 years ago

What happens when there is an element and there is another element what is that called?

Oksanka [162]

It is called Chemical compound

$\rightarrow{Element\:+\:Element\:=\:Chemical\: compound}$

$\\$

for example →

$\large\rightarrow\pink{HCL}$

$\large\rightarrow{\color{yellow}{{CO}_{2}}}$

$\large\rightarrow{\color{lime}{NaOH}}$

5 0

2 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

aleksandr82 [10.1K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 13.83 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.300 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.300M=\frac{\text{Moles of NaI}}{0.200L}\\\\\text{Moles of NaI}=(0.300mol/L\times 0.200L)=0.06mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.06 moles of NaI will produce = $\frac{1}{2}\times 0.06=0.03mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.03 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.03mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.03mol\times 461.1g/mol)=13.83g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 13.83 grams

6 0

3 years ago

If an element has an atomic number of 11 and a mass number of 23, how many protons, neutrons, and electrons does it have

Lostsunrise [7]

Protons = atomic number = 11
electrons = proton number = 11
neutrons = mass number - atomic number = 23-11 = 12

5 0

3 years ago