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IrinaK [193]
3 years ago
12

In order to murder somebody in the bathtub, police say that the suspect would need at least 50,000 grams of water. Was there eno

ugh water? Can Steve be the murderer? 2 H2 + O2 -------> 2H20 Police said that there were 5,000 grams of H2 in the room. Convert grams of hydrogen into grams of water and see if there was enough.​
Chemistry
1 answer:
Gre4nikov [31]3 years ago
8 0

Answer:

um, not sure how to answer this haha

Explanation:

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The molecular mass of air, at standard pressure and temperature, is approximately 28.97 g/mol. Calculate the mass of 3.33 moles
hoa [83]
Note: You are calculating mass which is determine the gram(g)
You will have to cancel out the mol
(28.97 g/mol) * mol will give grams by itself
Given the mass 3.33 moles of air
28.97 g/mol * 3.33 mol = 96.47 grams
Solution: 96.5 grams
5 0
2 years ago
If you want to prepare 5 liters of a 0.35m solution of nh4cl, how many grams of salt
harina [27]
Answer is: mass fo ammonium chloride is 93.625 grams.
V(NH₄Cl) = 5 L.
c(NH₄Cl) = 0.35 M.
n(NH₄Cl) = V(NH₄Cl) · c(NH₄Cl).
n(NH₄Cl) = 5 L · 0.35 mol/L.
n(NH₄Cl) = 1.75 mol.
M(NH₄Cl) = 14 + 1·4 + 35.5 · g/mol = 53.5 g/mol.
m(NH₄Cl) = n(NH₄Cl) · M(NH₄Cl).
m(NH₄Cl) = 1.75 mol · 53.5 g/mol.
m(NH₄Cl) = 93.625 g.
5 0
3 years ago
Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell
Oxana [17]

Answer:

\mathbf {density \ d  =4.4845 \ g/cm^3}

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

\mathtt{ density \ d  = \dfrac{z \times molar\  mass  \ (M)}{ edge \ length \ (a)  \ \times avogadro's \ number \ (N)}}

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

\mathtt{ density \ d  = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}

\mathtt{ density \ d  = \dfrac{ 212.81}{47.4540533}}

\mathbf {density \ d  =4.4845 \ g/cm^3}

5 0
3 years ago
When 20.00 mL of an unknown monoprotic acid is titrated with 0.125 M NaOH, it takes 15.00 mL to reach the endpoint. What is the
ohaa [14]

Answer:

About 0.0940 M.

Explanation:

Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:


\displaystyle \text{HA}_\text{(aq)} + \text{OH}^-_\text{(aq)} \longrightarrow \text{H$_2$O}_\text{($\ell$)} + \text{A}^-_\text{(aq)}

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:

\displaystyle \begin{aligned} 15.00\text{ mL} &\cdot \frac{0.125\text{ mol NaOH}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} \\ \\  &\cdot \frac{1\text{ mol OH}^-}{1\text{ mol NaOH}} \cdot \frac{1\text{ mol HA}}{1\text{ mol OH}^-}\\ \\  & = 0.00188\text{ mol HA}\end{aligned}

Therefore, the molarity of the original solution was:


\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}

In conclusion, the molarity of the unknown acid is about 0.0940 M.

3 0
2 years ago
When Iron (III) Choride is added to a solution of synthesized aspirin what is the method of detection
zubka84 [21]

Answer:

seeing the color change

Explanation:

3 0
2 years ago
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