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Assoli18 [71]
3 years ago
5

A cyclist accelerates from 0 m/s to 10 m/s in 5 seconds. What is his acceleration ?

Chemistry
1 answer:
choli [55]3 years ago
7 0

Answer:

Vi=0

Vf=10m/s

t=5s

a=?

Explanation:

a=Vf-Vi/t

a=10-0/5

a=10/5

a=2m/s^2

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Mole Conversion: How many molecules are there in 6.70 moles of methane, CH4 ​
raketka [301]

Answer:

Explanation:

1 mol of methane = 6.02 * 10^23 molecules

6.70 mol of methane = x

Cross multiply

x = 6.70 * 6.02 * 10^23

x = 4.033 * 10^23 molecules.

8 0
3 years ago
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To calculate the percentage of people who have a particular allele population studies are conducted to collect data.true or fals
kondaur [170]
I think is is going to be truth because
6 0
3 years ago
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

8 0
4 years ago
The volume of a sample of oxygen is 300.0 mL, when the pressure is 1.00 atm and the temperature is 300K. At what pressure will t
Brut [27]

Answer:

The new pressure is 0.5 atm

Explanation:

Step 1: Data given

Volume of oxygen = 300 mL = 0.300 L

Pressure = 1.00 atm

Temperature = 300 K

The volume increases to 1000mL = 1.00 L

The temperature increases to 500 K

Step 2: Calculate the new pressure

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 = the initial pressure = 1.00 atm

⇒with V1 = the initial volume = 0.300 L

⇒with T1 = the initial temperature = 300 K

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the increased volume = 1.00 L

⇒with T2 = the increased temperature = 500 K

(1.00 atm* 0.300 L)/300 K = (P2 * 1.00L) / 500 K

P2 = (1.00 *0.300 * 500) / (300 *1.00)

P2 = 0.5 atm

The new pressure is 0.5 atm

4 0
3 years ago
A precipitation reaction is caused by mixing 100. mL of 0.25 M K2Cr2O7 solution with 100. mL of 0.25 M Pb(NO3)2 solution. When t
Sergeeva-Olga [200]

Answer:

Neither is affected

Explanation:

The reaction occurs as follows:

K₂Cr₂O₇ + Pb(NO₃)₂ → PbCr₂O₇ + 2K⁺ + 2NO₃⁻

That means per mole of reaction you will have two moles of both K⁺ and NO₃⁻.

But volume is also doubled, doing that concentration of spectator ions doesn't change.

Right answer: Neither is affected

I hope it helps!

8 0
3 years ago
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