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Advocard [28]
2 years ago
11

The geometry of a ______________ makes it a very stable shape.

Chemistry
1 answer:
boyakko [2]2 years ago
5 0

Answer:

triangles

Explanation:

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You have 1.88 moles of NaOH and an excess of H2SO4. How many grams of H2O will be produced?
KATRIN_1 [288]

Answer:

2 NaOH + H2SO4 2 H2O + Na2SO4

How many grams of sodium sulfate will be formed if you start with 200.0

grams of sodium hydroxide and you have an excess of sulfuric acid?

355.3 grams of Na2SO4

200.0 g NaOH 1 mol NaOH 1 mol Na2SO4 142.1 g Na2SO4

40.00 g NaOH 2 mol NaOH 1 mol Na2SO4

= 355.3 g

Explanation:

3 0
2 years ago
ALL I ask is you answer correctly I'm having a bad day ! :(
Lilit [14]

Answer:

i have no clue whats going on here but imma act like i do.....

Explanation:

7 0
2 years ago
Convert 1.50μm2 to square meters
IrinaVladis [17]
1 micro meter = 10^{-6}  meters
1 μm ^2 = 1 μm*1μm = 10^{-6} *10^{-6} =  10^{-12} meters

1.5 μm^2 = 1.5 * 10^{-12} meters

μm^2 is a unit for surface. First you want to convert μm to meters which is unit for length and if you multiply units for length you get unit for surface.
4 0
2 years ago
LAB: predicting products
kondor19780726 [428]

Answer:

1) synthesis MgI2

2) double replacement CuS + (HCl)2

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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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