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REY [17]
3 years ago
14

youare given the solution of l!ad nitrate in order to obtaina yellow precipite you should mix with it if a solution a)sulphric a

cid b)citric acid c)carbonic acid d) hydrochloric acid​
Chemistry
1 answer:
Fiesta28 [93]3 years ago
7 0

Answer:

B. citric acid it just makes sense

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WILL GIVE BRAINLIEST
Marianna [84]
Answer:

B, C, and D

Explanation:

A is the only one in which two components are being combined. The point is to separate the mixture, so that is the only one that would not apply.
7 0
3 years ago
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A student is studying asbestos, a mineral that causes cancer in humans. One of the components of asbestos is silicon, a somewhat
kondaur [170]

Answer:

The characteristic of silicon that is most closely related to its chemical reactivity is that;

It is found in nature in mainly as oxides and silicates

Explanation:

Silicon, which is a member of group 14 of the periodic table has the electron configuration of [Ne]3s²3p² and has a high affinity for oxygen such that the the oxides are known as silicates and it is almost impossible to find pure silicon in nature and it is found in the universe as silica and silicates

Silicate minerals make up over 90% of the Earth's crust such that, by mass silicon is only surpassed by oxygen as the most abundant element found in the Earth's crust.

6 0
3 years ago
820 g Li2SO4 is dissolved into 2500 mL of solution. What is the molar concentration?
Andreas93 [3]

Molarity is defined as the moles of solute per liter of solution. M=\frac{n}{V}. Where M is molarity, n is the number of moles and V is the volume. First we must find the molar mass of Li_2SO_4 which is 109.98 g/mol

Li_2SO_4= 2 \times Li_{Ar} + S_{Ar} + 4 \times O_{Ar}\\= (2 \times 6.941 + 32.1 + 4 \times 16.0) = 109.98 g/mol

820\ g\ Li_2SO_4 \times \frac{mol}{109.98\ g}= 7.46\ mol\ Li_2SO4

Then we find the molarity using above equation

M= \frac{n}{V} = \frac{7.46 mol}{2500ml} \times\frac{1000ml}{L} = 2.98\ M


6 0
3 years ago
Which best illustrates the way in which radiation? Giving brainy.
Ilya [14]

Answer:

I dont really understand your question. but maybe this might help.

https://quizlet.com/283377931/radiation-flash-cards/

Explanation:

4 0
2 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
2 years ago
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