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lina2011 [118]
3 years ago
8

B) If I start with 225.0 grams of lead (II) sulfite and 315.0 grams of sodium iodide, how many

Chemistry
1 answer:
Rama09 [41]3 years ago
8 0

Answer:My answer is in the photo

Explanation:

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A sample of nitrogen gas has the temperature drop from 250.°c to 150.°c at constant pressure. what is the final volume if the in
Leviafan [203]
Charles law gives the relationship between temperature of gas and volume of gas.
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where   V- volume , T - temperature and k - constant 
\frac{V1}{T1} =  \frac{V2}{T2}
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,

\frac{310 mL}{523 K} =  \frac{V}{423K}
V = 251 mL
the new volume is 251 mL 
6 0
3 years ago
What happens during meiosis that does NOT happen during mitosis?
Furkat [3]

Answer:

B. two rounds of cell division

4 0
2 years ago
Read 2 more answers
In the Bronsted-Lowry model of acids and bases, an _____ is a hydrogen donor and a _____ is a hydrogen acceptor.
Natasha_Volkova [10]

Answer:

According to Bronsted-lowry concept  an acid is a hydrogen donnor and a base is a hydrogen acceptor.

Explanation:

4 0
3 years ago
How many moles of h2o will be produced from 2.9 g of HCl reacting with Ca(OH)2?
Igoryamba

Answer:

0.080 mol

Explanation:

 M(HCl) = (1.0 +35.5) g/mol = 36.5 g/mol

2.9g*1mol/36.5 g = 0.0795 mol HCl

                              Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

from reaction                            2 mol                   2 mol

given                                       0.0795 mol           x mol

x = 0.0795 mol ≈0.080 mol

7 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

7 0
3 years ago
Read 2 more answers
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