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fomenos
3 years ago
10

Why do atoms receive a positive or negative charge when they give or take electrons? Include the nucleus in your

Chemistry
1 answer:
marysya [2.9K]3 years ago
4 0
Atoms receive a positive charge when they take electrons. Atoms give of negative electrons.
You might be interested in
What kind of electrons are involved in chemical bonding?
mamaluj [8]
Bonding electrons are involved in chemical bonding these electrons have their valnce shell incomplete
6 0
3 years ago
What is the density of a sample of a substance with a volume of 120ml<br> and a mass of 90g?
tigry1 [53]

Answer:

<h3>The answer is 0.75 g/mL</h3>

Explanation:

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

density =  \frac{mass}{volume}  \\

From the question

mass = 90 g

volume = 120 mL

We have

density =  \frac{90}{120}  =  \frac{9}{12}  =  \frac{3}{4}  \\

We have the final answer as

<h3>0.75 g/mL</h3>

Hope this helps you

6 0
3 years ago
Select the correct answer.
timofeeve [1]

Answer:

D

Explanation:

4 0
3 years ago
Read 2 more answers
You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
miss Akunina [59]

Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

5 0
3 years ago
How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!
NNADVOKAT [17]

Answer:

Approximately 1.53 \times 10^{23} formula units (0.254\; \rm mol).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium (\rm Mg) and chlorine (\rm Cl):

  • \rm Mg: 24.305.
  • \rm Cl: 35.45.

In other words, the mass of 1\; \rm mol of \rm Mg atoms would be (approximately) 24.305\; \rm g.

Likewise, the mass of 1\; \rm mol of \rm Cl atoms would be approximately 35.45\; \rm g.

One formula unit of the ionic compound \rm MgCl_{2} includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of 1\; \rm mol of the formula units of this compound.

The formula \rm MgCl_{2} includes one \rm Mg atom and two \rm Cl atoms.

Hence, every formula unit of \rm MgCl_{2} \! would include the same number of atoms: one \rm Mg\! atom and two \rm Cl\! atoms. There would be 1\; \rm mol of \rm Mg atoms and 2\; \rm mol of \rm Cl atoms in 1\; \rm mol\! of \rm MgCl_{2} formula units.

Thus, the mass of 1\; \rm mol\! of \rm MgCl_{2} formula units would be equal to the mass of 1\; \rm mol of \rm Mg atoms plus the mass of 2\; \rm mol of \rm Cl atoms. (The mass of 1\; \rm mol\!\! of each atom could be found from the relative atomic mass of each element.)

\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}.

In other words, the formula mass of \rm MgCl_{2} is 95.205\; \rm g \cdot mol^{-1}.

Therefore, the number of formula units in m = 24.2\; \rm g of \rm MgCl_{2} would be:

\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}.

Multiple n by Avogadro's Number N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1} to estimate the number of formula units in 0.254\; \rm mol:

\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}.

6 0
3 years ago
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