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vladimir1956 [14]
3 years ago
11

What do you think the volume of a perscription bottle would be? plz help me

Chemistry
1 answer:
noname [10]3 years ago
5 0
The bottle is 200 ml
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Which is one of the following is a nonmetal
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Br is a non-metal. The other answers like Os, Ir, W are not combinations while Sr is a metal. So Br is your answer.
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3 years ago
Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the
kari74 [83]

Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH  = 3.40+ log {0.170 /0.105}

pH =  3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol  = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

4 0
2 years ago
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Serial dilution problem: Six test tubes are placed in a rack. To each tube add 4 mL of saline solution. Now to the first tube ad
ArbitrLikvidat [17]

Answer:

The dilution factor of protein in tube # 4 is 125. Molar concentration is 0.0088 M protein

Explanation:

The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:

Initial volume * initial concentration = final volume * final concentration

0.5 mL * 10M = 5mL * final concentration

1.1 M = final concentration = main solution concentration

Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.

Dilution factor = Main solution concentration/tube 4 concentration

Dilution factor = 1.1/0.0088 = 125

I hope my answer helps you

3 0
3 years ago
How many particles are there in 5.1 moles of NH4S02?
miskamm [114]

Answer:

12 to 24

Explanation:

i learned this

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What does matter mean?
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Anything that has mass and occupies space.
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