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3 years ago

I really need an answer please help

Chemistry

2 answers:

saul85 [17]3 years ago

7 0

The answer is A because at this moment the ball has the greatest height from the ground compared to the other locations.

VladimirAG [237]3 years ago

3 0

Answer:

56g the time to convey he must take care of the full stops and then try to combine the two sentences that the time to convey

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Glucose, a common carbohydrate, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a

lesya [120]

The formation of carbondioxide is given by equation as

C + O2 ----> CO2(g)

As per balanced equation one mole of carbondioxide is obtained from one mole of carbon

if 44 g of CO2 is formed it means it contains 12 g of carbon

if 1g of CO2 is formed it means it contains 12/44 g of carbon

if 0.528g of CO2 is formed it means it contains = 12 X 0.528 / 44 g of carbon

= 0.144g of carbon

Similarly for water

H2 + 0.5O2 ---> H2O

so one mole of water is obtained from one mole of H2

if 18g of water is formed it means 2g H2 is present

if 1g of water is formed it means 2/18 g of H2 is present

if 0.216g of water if formed it means 2 X 0.216 / 18g of H2 is present = 0.024g of Hydrogen

In glucose we have Carbon , Hydrogen and oxygen

As given the total mass of sample = 0.360g

Mass of carbon = 0.144 , moles of carbon = 0.144 / 12 = 0.012

Mass of hydrogen = 0.024 , moles of hydrogen = 0.024 /1 = 0.024

mass of oxygen = 0.360 - (0.144+0.024) = 0.192

Moles of oxygen = 0.192 / 16 = 0.012

So mole ratio of C : H : O = 0.012 : 0.024 : 0.012 = 1:2:1

Empirical formula = CH2O

7 0

4 years ago

Americium-241 is used in smoke detectors. it has a first order rate constant for radioactive decay of k=1.6×10−3yr−1. by contras

Helga [31]

This question is missing the part that actually asks the question. The questions that are asked are as follows:

(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.

(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.

We can use the equation for a first order rate law to find the amount of material remaining after 4 days:

[A] = [A]₀e^(-kt)

[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.

(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.

4 days x 1 year/365 days = 0.0110

A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg

The decay of americium is so slow that no noticeable change occurs over 4 days.

(b) We can simply plug in the information of iodine-125 and solve for A:

A = (1.00)e^(-0.011 x 4)
A = 0.957 mg

Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.

7 0

4 years ago

The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th

Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_2$ = rate of reaction at $T_2$

$K_1$ = rate of reaction at $T_1$

$Ea$ = activation energy of the reaction

R = gas constant = 8.314 J/K mol

$E_a=300 kJ/mol=300,000 J/mol$

$T_2=625^oC=898.15 K,T_1=525^oC=798.15 K$

$\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]$

$\log (\frac{K_2}{K_1})=2.185666$

$K_2=153.344\times K_1$

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0

3 years ago

An objects volume can be found by adding its mass by its?

Inessa05 [86]

Volume* density=MASS

4 0

4 years ago

how many moles of sodium hydroxide would have to be added to 250 ml of a 0.403 m acetic acid solution, in order to prepare a buf

jonny [76]

The volume of the buffer solution having a ph value is calculated by henderson's hasselbalch equation.

Buffer solution is water based solution which consists of a mixture containing a weak acid and a conjugate base of the weak acid. or a weak base and conjugate acid of a weak base.it is a mixture of weak acid and a base. The pH of the buffer solution is determined by the expression of the henderson hasselbalch equation.

pH=pKa + log [salt]/[acid]

Where, pKa =dissociation constant , A- = concentration of the conjugate base, [HA]= concentration of the acid. Here, a buffer solution contains 0.403m acetic acid and 250 ml is added in order to prepare a buffer with a ph of 4.750. Putting all the values in the henderson hasselbalch equation we find the pH of the buffer solution.

To learn more about hendersons hasselbalch equation please visit:

brainly.com/question/13423434

#SPJ4

6 0

1 year ago