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wariber [46]
3 years ago
10

CAN SOMEONE go through this with me pleaseeeSTEP BY STEP AND TELL ME THE ANSWERS!!!thx

Chemistry
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

the answer would be runner A ,because they were ahead of all the other runners and runner A got there before the other runner so runner A has the fastest average speed

Explanation:

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A gas occupies a volume of 67.54 liters at 158°C and 4.87 atm pressure. Calculate the number of moles of this gas.
Katarina [22]

Answer:

The number of moles of the gas is 9.295 moles or 9.30 moles

Explanation:

We use PV = nRT

Where P = 4.87 atm;

V = 67.54 L

R= 0.0821Latm/molK

T = 158 C = 158 +273 K = 431 K

the number of moles can be obtained by substituting the values in the respective columns and solve for n

n = PV / RT

n = 4.87 * 67.54 / 0.0821 * 431

n = 328.9198 / 35.3851

n = 9.295moles

The number of moles is approximately 9.30moles.

6 0
3 years ago
Identify the substance that is oxidized. Copy the oxidation half reaction for this substance.
slava [35]

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4 0
3 years ago
How does fabric softener affect the flammability of different fabrics?
Karolina [17]
<span>Research shows that fabric softeners decrease flame resistance. The flame resistance of a fabric can be increased in a variety of ways. Most often, the fabric is treated with special flame resistant chemicals. Synthetic fabrics, such as polyester, are naturally flame resistant. These fabrics have a property called thermoplasticity, where the fabric naturally melts and shrinks when exposed to flame. This prevents these fabrics from igniting and burning as easily as other fabrics. Basic properties of a fabric, such as porosity, can also influence flame resistance. Larger pores allow for oxygen to be more present in the fabric, making it more flammable.</span>
5 0
3 years ago
Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0
3 years ago
How many grams are in 0.02 moles of beryllium iodide, Bel2?
Flura [38]
This set up of a conversion table should show you that if you multiply the grams of BeI2 times .02 moles, it equals <span>5.256 g (your answer) </span>

8 0
3 years ago
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