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Eduardwww [97]
3 years ago
9

How many significant figures are in the measurement 40,500 mg?

Chemistry
1 answer:
schepotkina [342]3 years ago
3 0

b) three

this is because all integers are sig figs, and all numbers between integers are sig figs. This makes the 40,5 part of 40,500 significant. Place holder zeroes that are not after a decimal are not significant, so the last two zeroes of the number are not significant.

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To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?
poizon [28]
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

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7 0
3 years ago
3-chloropropylamine what is?
marta [7]

Answer:

a chemical compound

6 0
2 years ago
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
Valentin [98]

Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

Best regards.

8 0
3 years ago
Read 2 more answers
You are asked to prepare 500. mL 0.150 M acetate buffer at pH 5.10 using only pure acetic acid ( MW = 60.05 g/mol, p K a = 4.76
Amanda [17]

Answer:

4.504g of acetic acid

Explanation:

The acetic acid in reaction with NaOH produce acetate ion, thus:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

<em>That means the moles of acetate buffer comes, in the first, from the acetic acid</em>

As you need 500mL (0,500L) of a 0.150M acetate buffer, moles are:

0.500L × (0.150mol / 1L) = <em>0.075 moles of acetate</em>. That is:

0.075mol = [CH₃COO⁻] + [CH₃COOH]

Thus, grams of acetic acid you need to prepare the buffer are:

0.075 moles acetic acid × (60.05g / 1mol) = <em>4.504g of acetic acid</em>

6 0
3 years ago
Give two examples of electrical energy being transformed in to another type of energy.Tell what energy is produced
seraphim [82]

Answer:

1.Electric generators (mechanical energy to electricity)

2.Windmills (mechanical energy to electricity)

7 0
2 years ago
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