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Damm [24]
3 years ago
13

A piper delivers 9.98 g of water at 19 degrees Celsius. What does the pipet deliver

Chemistry
2 answers:
leva [86]3 years ago
5 0

Answer:

the volume = 9.99 ml

Explanation:

in your Q we need to calculate the volume does the pipet deliver so , we are going to use this formula:

the volume = mass / density

here we need to know the density of water at a certain temperature 19 degrees Celsius ,so I used an external source   to get the density of water at 19 degrees Celsius because it is changing with different temperatures

where mass here = 9.98 g

and  the density of water at  temperature 19 degrees Celsius=  0.998405mL

-1

by substitution :

the volume = 9.98 g / 0.998405mL

-1

                   ≅ 9.996 ml = 10 ml

yuradex [85]3 years ago
5 0

Answer:

9.99 ml water will be delivered from pipet

Explanation:

When Water is being delivered by the pipet the temperature is 19 degree Celsius. Also know that dependence of density is on temperature although this temperature dependence is inverse in nature that is when temperature increases density decreases and vice versa.

Now we also know the density of water at 20 ° Is 0.99 gml-1.

At this temperature 9.98 gram will occupy a volume= 0.98/0.998= 9.997ml

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In a first order reaction 40% of reactant gets converted into product in 30 minutes. What time would it require to convert 75% i
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56 minutes and 15 seconds I believe
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3 years ago
How many grams of hcl can be produced from the reaction of 23 moles of h2 and 875g of cl2?
max2010maxim [7]

Answer: 899 g

Explanation:

Since the mass of a mole of diatomic chlorine is about 70.9 g/mol, this means that 875 grams is about 875/70.9 = 12.3 mol.

Hence, we know that chlorine is the limiting reactant, and that about 12.3 moles of chlorine will be consumed.

This means that 12.3(2)=24.6 grams of HCl will be produced, which will have a mass of about (24.6)(36.46)=899 g

8 0
2 years ago
If the outside of your flask is not dry when the first mass determination is made and it is dried for the second mass determinat
svlad2 [7]

Answer:

You'll experience a grater deviation

Explanation:

<em>You'll experience a greater deviation in your measurements, meaning your measures will have a bigger difference between them, and the greater these deviations the less accurate will be the measuring.</em> This happens mainly because you're not replicating the measurement with the exact same conditions, in one of them you'll have an extra mass from the water.

I hope you find this information useful and interesting! Good luck!

8 0
3 years ago
What us the % composition of oxygen (O) in iron (III) hydroxide, Fe(OH)<br> 3?
S_A_V [24]

Answer:

44.91% of Oxygen in Iron (III) hydroxide

Explanation:

To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:

<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>

<em />

<em>Molar mass Fe(OH)3 and oxygen:</em>

1Fe = 55.845g/mol*1 = 55.845

3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen

3H = 1.008g/mol*3 = 3.024

55.845 + 48.00 + 3.024 =

106.869g/mol is molar mass of Fe(OH)3

% Composition of oxygen is:

48.00g/mol / 106.869g/mol * 100 =

<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
7 0
3 years ago
Calculate the amount of heat necessary to raise the temperature of 135.0 g of water from 50.4°F to 85.0°F. The specific heat of
MAXImum [283]

Here we have to calculate the heat required to raise the temperature of water from 85.0 ⁰F to 50.4 ⁰F.

10.857 kJ heat will be needed to raise the temperature from 50.4 ⁰F to 85.0 ⁰F

The amount of heat required to raise the temperature can be obtained from the equation H = m×s×(t₂-t₁).

Where H = Heat, s  =specific gravity = 4.184 J/g.⁰C, m = mass = 135.0 g, t₁ (initial temperature) = 50.4 ⁰F or 10.222 ⁰C and t₂ (final temperature) = 85.0⁰F or 29.444 ⁰C.

On plugging the values we get:

H = 135.0 g × 4.184 J/g.⁰C×(29.444 - 10.222) ⁰C

Or, H = 10857.354 J or 10.857 kJ.

Thus 10857.354 J or 10.857 kJ heat will be needed to raise the temperature.

6 0
3 years ago
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