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3 years ago

URGENT!!!! Please respond in less than 5 mins hurry!!!!?!!?!!??!?

Chemistry

1 answer:

AleksAgata [21]3 years ago

5 0

Answer:

1. A

2. B

Explanation:

1. Most of the answers don't make that much sense, so using process of elimination I resulted in this answer. I recommend you also try to check with others.

2. This question also was pretty confusing but when using process of elimination I resulted in either A or B. To me B makes more sense.

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What mass of solute is contained in 25.4 ml of a 1.56 m potassium bromide solution?

mash [69]

Answer is: mass of <span>potassium bromide is 4.71 grams.
V(KBr) = 25.4 mL </span>÷ 1000 mL/L = 0.0254 L, volume of solution.
c(KBr) = 1.56 mol/L.
n(KBr) = c(KBr) · V(KBr).
n(KBr) = 1.56 mol/L 0.054 L.
n(KBr) = 0.0396 mol, amount of substance.
m(KBr) = n(KBr) · M(KBr).
m(KBr) = 0.0396 mol · 119 g/mol.
m(KBr) = 4.71 g.
M - molar mass.

3 0

3 years ago

Read 2 more answers

What substance do you place in the burette during a titration

strojnjashka [21]

Answer:

Acid solution

Explanation:

In acid-base titrations carried out in school and college labs, many of the older generation, well-informed teachers told their students that the acid solution should be taken in the burette.

5 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

How many molecules are in 8.3 moles?

devlian [24]

1 mole=6.02 ×10

8.3 moles=?

8.3×6.02×10=499.66 molecules

therefore ghe no. of molecules = 499.66

6 0

3 years ago

the mass of chemical a is 2g and the mass of chemical b is 5g. if the two chemical are mixed and a chemical reaction takes place

OverLord2011 [107]

Answer:

the answer is 7g

Explanation:

I'm not really sure about the answer but the way i did it is i added 2+5g

5 0

2 years ago