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schepotkina [342]
3 years ago
13

How many grams are in 4.5 moles NaF (show work ) ^ill mark u as brainlister please

Chemistry
2 answers:
Aleks04 [339]3 years ago
8 0

Answer:

190 grams

Explanation:

Find the molar mass of NaF

Na = 23.0

F = 19.0

23 + 19 = 42

Now use this dimensional analysis to show your work

4.5 mol NaF * ( 42.0 grams of NaF / 1 mol NaF)

Moles cancel out.

4.5 x 42.0 = 189

With sig figs, it is 190.

So the final answer is 190 grams in 4.5 moles NaF

Anton [14]3 years ago
3 0

Answer:

There are 188.96 gm of moles in NaF

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The space between corresponding points (adjacent crests) in adjacent cycles is known as the wavelength () of a waveform signal that is sent in space or down a wire.

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Putting the values in the formula

λ = (6.626 x 10⁻³⁴)(2.9 × 10⁸) / (3.65 x 10⁻¹⁹) = 5.42 x 10⁻⁷ m.

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2 years ago
Need help what is the answer
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6 0
2 years ago
For the reaction H2PO4- HAsO4 2-HPO4 2-+H2AsO4- <br> what species are a conjugate acid-base pair?
aleksklad [387]

Ionic equation:

\text{H}_2\text{PO}_4^{-} + \text{HAsO}_4^{2-} \to \text{HPO}_4^{2-} + \text{H}_2\text{AsO}_4^{-}

The acid and base in a conjugate pair differ by only one proton \text{H}^{+}. The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.

\text{H}_2\text{PO}_4^{-} loses one proton to produce \text{HPO}_4^{2-} in this reaction.

\text{H}_2\text{PO}_4^{-} \to \text{H}^{+} + \text{HPO}_4^{2-}

Meanwhile, \text{HAsO}_4^{2-} gains one proton to form \text{H}_2\text{AsO}_4^{-}.

\text{HAsO}_4^{2-} + \text{H}^{+} \to \text{H}_2\text{AsO}_4^{-}

Therefore

  • \text{H}_2\text{PO}_4^{-} is the conjugate acid  \text{HPO}_4^{2-}, its conjugate base.
  • \text{HAsO}_4^{2-} is the conjugate base of \text{H}_2\text{AsO}_4^{-}, its conjugate acid.
8 0
3 years ago
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Citrus2011 [14]

Answer:

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4 0
3 years ago
Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen
olga55 [171]

Answer: Rate law=k[A]^1[B]^2, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is 3L^2mol^{-2}s^{-1}

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Rate=k[A]^x[B]^y

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: 1.2\times 10^{-2}=k[0.10]^x[0.20]^y    (1)

From trial 2: 4.8\times 10^{-2}=k[0.10]^x[0.40]^y    (2)

Dividing 2 by 1 :\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}

4=2^y,2^2=2^y therefore y=2.

b) From trial 2: 4.8\times 10^{-2}=k[0.10]^x[0.40]^y    (3)

From trial 3: 9.6\times 10^{-2}=k[0.20]^x[0.40]^y   (4)

Dividing 4 by 3:\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}

2=2^x,2=2^1, x=1

Thus rate law is Rate=k[A]^1[B]^2

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1:  1.2\times 10^{-2}=k[0.10]^1[0.20]^2

k=3 L^2mol^{-2}s^{-1}.



6 0
3 years ago
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