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Alla [95]
3 years ago
13

How many grams of H3PO4 are in 234 mL of a 2.5 M solution of H3PO4?

Chemistry
1 answer:
motikmotik3 years ago
5 0

Answer:

57 grams of H3PO4

Explanation:

M= moles/ liters

convert mL to L

234 mL x 1L/1000mL = 0.234L

Rearrange the Molarity formula to solve for moles.

moles= MxL

moles= 2.5M x 0.234L

moles= 0.585 mol

Use the molar mass of H3PO4 to get to grams

0.585 mol x 97.994 grams/1 mol = 57.326 grams of H3PO4

round to two sig figs for 57 grams

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What is the rate law for the reaction 2A + 2B + 2C --> products
-Dominant- [34]

Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0
4 years ago
Calculate the ionization constant for the following acids or bases from the ionization constant of its conjugate base or conjuga
suter [353]

Answer:

7.41 × 10⁻⁵

Explanation:

Let's consider the basic dissociation reaction of trimethylamine (CH₃)N).

(CH₃)N + H₂O = (CH₃)NH⁺ + OH⁻

According to Brönsted-Lowry, in this reaction (CH₃)N is a base and (CH₃)NH⁺ is its conjugate acid. The pKb for (CH₃)N is 9.87. We can calculate the pKa of (CH₃)NH⁺ using the following expression.

pKa + pKb = 14

pKa = 14 - pKb = 14 - 9.87 = 4.13

Then, we can calculate the acid dissociation constant for (CH₃)NH⁺ using the following expression.

pKa = -log Ka

Ka = antilog - pKa = antilog -4.13 = 7.41 × 10⁻⁵

8 0
3 years ago
Which composition of water moves to begin a deep water current
jasenka [17]
Cold and high salinity. Hope this helps :)
4 0
3 years ago
Read 2 more answers
Not all elements readily fit in the metal or non metal groupings. These elements are called metalloids. Examine the table below
crimeas [40]
Brown luster moderately high conductivity
3 0
3 years ago
100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of vo
gtnhenbr [62]

Answer:

8.912x10^-18

Explanation:

-dn/dt = pANa/2piMRT

100 g = initial copper

Number of moles = 100/63.546

= 1.5736

Mass of copper left = 100-10.0168

= 89.9832

Moles = 89.9832/63.546

= 1.4160

dn = 1.4160-1.5736

= -0.1576

dt = 2 hrs

A = 3.23mm² = 3.23x10^-6

M = 63.546

T = 0.0821

T = 1508k

Na = 6.023x10²³

When we insert all these into the formula above

We get

P = 8.912x10^-18atm

4 0
3 years ago
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