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3 years ago

Describe each type of matter.

Chemistry

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer:

there all made of atoms

Explanation:

you did not specify which types of matter

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Write the reaction when solid lead(ii) nitrate is put into water:

Whitepunk [10]

Solid lead nitrate in water gives lead oxide and nitric acid

Pb(NO3)2 + H2O ---> PbO + 2 HNO3

7 0

3 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago

Density is the ratio of mass to volume. Another way to say this is

Oliga [24]

Answer:

my teacher would let u say the density to mass is the ratio but i think everyone's different

5 0

3 years ago

Element m reacts with fluorine to form an ionic compound with the formula mf4. the m cation has 18 electrons. element m is:____.

polet [3.4K]

Element M reacts with fluorine to form an ionic compound with the formula MF₄. The M cation has 18 electrons. Element M is Sn.

<h3>What is Ionic Compound ?</h3>

An ionic compound is defined as the compound made up of ions that formed charge particles when an atom loses or gains electrons.

<h3>What is Cation ?</h3>

Cations are positively charged ions. It forms when an element lose one or more electrons.

The cation which has 18 electrons is Sn. Sn react with fluorine to form an ionic compound with the formula SnF₄ because Sn is large and F is very small in size.

Thus from the above conclusion we can say that Element M reacts with fluorine to form an ionic compound with the formula MF₄. The M cation has 18 electrons. Element M is Sn.

Learn more about the Ionic Compound here: brainly.com/question/2687188

#SPJ1

6 0

3 years ago

What is the molarity of NO−3 in 0.160M KNO3?

zubka84 [21]

A solution of KNO3 consists of ions of potassium and nitrate. The ionic equation is expressed as:

KNO3 = K+ + NO3-

There is 1 is to 1 ratio between the substances. So, the molarity of NO3- in the solution is calculated as follows:

0.160 mol / L KNO3 ( 1 mol NO3- / 1 mol KNO3 ) = 0.160 M NO3-

5 0

3 years ago