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natta225 [31]
3 years ago
9

How many molecules are in 88g of carbon dioxide?

Chemistry
2 answers:
Volgvan3 years ago
8 0
No of moles= 88/44 =2
therefore no. of molecules =

therefore no.of molecules= 12.046*10^23
DanielleElmas [232]3 years ago
3 0

Answer:there are 44 grams

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In which medium does sound travel the slowest?
ZanzabumX [31]
Sound actually travels slower in air.This is because sound is kinetic energy and has to pass from molecule to molecule.In gas the molecules are farther apart taking more time for it to pass,in water the molecules are closer so it takes less time for the sound to get from one molecule to another and with cast iron (since it's very dense) the molecules are very close allowing sound to travel quickly with ease.
5 0
3 years ago
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A sample of gas occupies 10 L at STP. What
puteri [66]

Pressure is 5.7 atm

<u>Explanation:</u>

P1 = Standard pressure = 1 atm

P2 = ?  

V1 = Volume = 10L

V2= 2.4L

T1 = 0°C + 273 K = 273 K

T2 = 100°C + 273 K = 373 K

We have to find the pressure of the gas, by using the gas formula as,

$\frac{P 1 V 1}{T 1}=\frac{P 2 V 2}{T 2}

P2 can be found by rewriting the above expression as,

$P 2=\frac{P 1 \times V 1 \times T 2}{T 1 \times V 2}

Plugin the above values as,

$P 2=\frac{1 \text {atm} \times 10 L \times 373 \mathrm{K}}{273 \mathrm{K} \times 2.4 \mathrm{L}}=5.7 \text { atm }

4 0
3 years ago
Muscles cells are called?
evablogger [386]

Answer:

They are called Myocyte.

Explanation:

A myocyte is said to be the type of cells that is found in muscle tissue. They are structurally long and also posses tubular cells that inhibits develop from myoblasts to form myogenesis process which are muscles.

Within a skeletal muscle fibre, there are many myocytes. arranged parallel to one another which gives the fibre a striated appearance. They are contractile parts of the muscle fibre.

Also, different forms of these myocytes has different qualities in our body muscles found in the cardiac, skeletal, and smooth muscle cells.

6 0
3 years ago
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Suppose there is 1. 00 l of an aqueous buffer containing 60. 0 mmol of acetic acid (pa=4. 76) and 40. 0 mmol of acetate. Calcula
8_murik_8 [283]

For 1. 00 l of an aqueous buffer containing 60. 0 mmol of acetic acid (pa=4. 76) and 40. 0 mmol of acetate, the pH of this buffer is 4.58.

<h3>What is handerson Hasselbalch equation? </h3>

It is expressed as:

pH = pKa + log [A-]/[HA]

where,

[A-] is the molar concentration of conjugate base

[HA] is the molar concentration of weak acid

Given,

pKa = 4.76

<h3>Calculation of concentration:</h3>
  • For acetic acid

C = n/V

= 60/1000

= 0.06M

  • For acetate

C = 40/1000

= 0.04M

Now, substituting values in equation we get,

pH = 4.76 + log(0.04/0.06)

pH = 4.76 + (-0.176)

pH = 4.58

Thus by using Henderson Hasselbalch equation we find the value of pH of the buffer is 4.58.

learn more about pH :

brainly.com/question/9529394

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6 0
1 year ago
A second- order reaction of the type A + B --&gt;P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0
andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = -\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B

If C_{A0} and C_{B0} are the inital concentrations and x the concentration reacted at time t, so C_A=C_{A0} -x and C_B=C_{B0} -x and the rate at time t is written as:

-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)

Separating variables and integrating

\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt

The integral in left side is solved by partial fractions, it can be used integral tables

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt

Using logarithm properties (ln x - ln y = ln(x/y))

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a C_{B0} of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, C_{A0}=0.075, C_{B0}=0.05, C_{B}=0.02, it implies that the quantity reacted, x, is 0.03 and C_{A}=0.075-x = 0.045. Then, the value of k would be

kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})

k = 16.21 \frac{dm^3}{mol*1h}

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt

but suppossing that C_A= C_{A0}/2=0.0375 so  C_B=C_{B0}- C_{A0}/2=0.0125, k=16.2 and the same initial concentrations. Replacing in the equation

t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})

t=1.71 h = 1.71*3600 s = 6.1*10^3 s  

For reactant B, C_B= C_{B0}/2=0.025 so  C_A=C_{A0}- C_{B0}/2=0.05, k=16.2 and the same initial concentrations. Replacing in the equation

t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})

t=0.71 h = 0.7*3600 s = 2.5*10^3 s  

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0
2 years ago
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