What happened to the celery if you put in food color water

A balloon is floating around outside your window. The temperature outside is -1 ∘C , and the air pressure is 0.700 atm . Your ne

Vladimir79 [104]

Answer:

The volume inside the balloon is = 121 $m^{3}$

Explanation:

Temperature T = - 1 °c = 272 K

Pressure = 0.7 atm = 71 k pa

No. of moles = 3.8

Mass of the gas inside the volume = 3.8 × 4 = 15.2 kg

From ideal gas equation

P V = m R T

Put all the values in above formula we get

71 × V =15.2 × 2.077 × 272

V = 121 $m^{3}$

Therefore the volume inside the balloon is = 121 $m^{3}$

6 0

2 years ago

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hi guys im giving away brainliest answer away for a good cause who ever gives me the best rating for my profile picture gets bra

vova2212 [387]

Answer:

Ive seen that pfp on google BUT its aight ig

Explanation:

6 0

2 years ago

Combustion of hydrogen releases 142 j/g of hydrogen reacted. How many kj of energy are released by the combustion of 16.0 oz of

Andrej [43]

Given the mass of hydrogen = 16.0 oz

Converting 16.0 oz hydrogen to pounds (lb) using the conversion factor 1 lb = 16 oz:

$16.0 oz * \frac{1 lb}{16 oz} =1 lb$

Converting 16.0 lb to g using the conversion factors 1 kg = 2.2 lb, 1 kg = 1000 g:

$1lb * \frac{1kg}{2.2lb}*\frac{1000g}{1kg}= 454.5 g$

Heat of combustion of hydrogen = 142 J/g

Calculating the heat released when 16.0 oz is combusted:

$454.5g H_{2} * \frac{142 J}{g} *\frac{1 kJ}{1000J}=64.5kJ$

5 0

2 years ago

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To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o

Romashka-Z-Leto [24]

Answer:

a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

$\pi r^{2}h$ ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

$r=\frac{d}{2}$

$=\frac{3.36x10^{2}}{2}$

= 168 m

Volume =

$=\frac{22\times 168^{2}\times 21.80}{7}$

$=1.93\times 10^{6} m^{3}$

2.Convert ppm to g/m3 and Solve for mass of F-

$1ppm = 1g/m^{3}$

$0.9ppm = 0.9g/m^{3}$

Because both ppm and g/m3 are same quantity .

$g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}$

$0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}$

$mass\ of\ F- =1.740g$

mass of F- required = 1.740 g

3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

$2Na+F_{2}\rightarrow 2NaF$

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

$\frac {84}{38}\times 1.74$

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0

3 years ago

A solution is made by mixing 0.82 grams of sodium acetate, 1 mL of 12 M acetic acid (pKa=4.76) and water to give a final volume

kompoz [17]

Answer:

This solution acts as an efficient buffer

Explanation:

the pH of a buffer solution can be described like this: $pH=pKa+log\frac{[base]}{[acid]}$

[acid]=[acetic acid]= $1mL.\frac{12mol}{1000mL} . \frac{1}{1L} = 0.012M$

[base]=[sodium acetate]= $0.82g.\frac{1mol}{82g} .\frac{1}{1l} = 0.01M$

replacing, $pH=4.76+log\frac{0.01M}{0.012M} =4.84$

If we add an acid, pH will decrease a little bit and if we add a base, pH wil increase a little bit.

lets supose that we change the rate by increasing [base] to 0.1, then

$pH=4,76+log\frac{0.1}{0.012} = 5.68$

and now lets supose that we increase [acid] to 0.1 $pH=4.76+log\frac{0.01}{0.1} = 3.76$

Big changes in concentration of base or acid doesn´t produce big changes in pH, in that way the mix of sodium acetate with acetic acid is a good buffer solution.

4 0

2 years ago

What happened to the celery if you put in food color water ​

What happened to the celery if you put in food color water