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Ivanshal [37]
3 years ago
12

How could you ensure that all the spectroscopes used the same calibration line and equation? What would you need to be sure was

always the same? What could be changed without affecting the calibration?
Chemistry
1 answer:
NISA [10]3 years ago
8 0
People had asked this many times and that is why they came up with methods and standards that will answer these type of questions. You can look it up in the NIST or the  National Institute for Standards and Technology.
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lactic acid, which consists of C, H, and O, has long been thought to be responsible for muscle soreness following strenuous exer
Minchanka [31]

Answer:

C3H6O3

Explanation:

I got you babe

3 0
3 years ago
Which of the following is an example of a contact force?
Virty [35]
C. A person pushing a box across the floor
7 0
3 years ago
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DNA profiling is one of the best sources of evidence to identify the guilty party because ______.
Vika [28.1K]

Answer:

B

Explanation:

4 0
4 years ago
The following reaction was performed in a sealed vessel at 791 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at
OlgaM077 [116]

Answer:

4.31 × 10²

Explanation:

Equation of the reaction;

H_{2(g)} + I_{2(g)}     ⇌     2HI_{(g)

The ICE Table is shown as follows:

                            H_{2(g)}         +         I_{2(g)}        ⇌     2HI_{(g)

Initial                    3.10                     2.50                  0      

Change                 - x                       -x                     + 2x      

Equilibrium        (3.10 - x)                0.0800              2x

From I_{2(g)}   ;

We can see that 2.50 - x = 0.0800

So; we can solve for x;

x = 2.50 - 0.0800

x = 2.42

H_{2(g)}  which = (3.10 -x) will be :

= 3.10 - 2.42

= 0.68

2HI_{(g) = 2x

= 2 (2.42)

= 4.84

K_c = \frac{[HI]^2}{[H_2][I_2]}

K_c = \frac{(4.84)^2}{(0.68)(0.0800)}

K_c =\frac{23.4256}{0.0544}

K_c = 430.62

K_c ≅ 431

K_c = 4.31 × 10²

6 0
3 years ago
It is desired to make 1.00 liter of 6.00 M nitric acid from concentrated 16.00 M HNO3.A) How many moles of nitric acid are in 1.
natima [27]

Answer:

A) 6.00 mol.

B) 0.375 L or 375 mL

C) 6.00 M

Explanation:

Hello,

A) In this case, from the definition of molarity, we compute the moles for the given volume and concentration:

n=M*V=1.00L*6.00mol/L=6.00mol

B) In this case, from the stock solution, the required volume is:

V=\frac{6.00mol}{16.00mol/L}=0.375L

C) In this case, we apply the following formula for dilution process:

M_1V_1=M_2V_2

Thus, solving for the final molarity, we obtain:

M_2=\frac{M_1V_1}{V_2}=\frac{16.00M*0.375L}{1.00L}\\  \\M_2=6.00M

Regards.

5 0
3 years ago
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