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2 years ago

Do you really get 7 years of bad luck if you crack a mirror :(

Chemistry

1 answer:

Mazyrski [523]2 years ago

4 0

Answer:

nope its a myth don't worry :)

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At sea level water usually boils at 100. °C and 760. mmHg pressure. On Mt Whitney, the pressure is about 560. mmHg. At what temp

WITCHER [35]

Answer:

91.7°C

Explanation:

We suppose you have a formula to work from. However, that is not supplied with this problem statement, so we looked one up.

The formula in the attachment is supposed to have good accuracy in the temperature range of interest. It gives vapor pressure of water in kPa, not mmHg, so we needed the conversion for that, too.

560 mmHg corresponds to about 74.66 kPa. The attached "Buck equation" formula is used to find the corresponding temperature. The exponential equation could be solved algebraically using logarithms and the quadratic formula, but we choose to find the solution graphically.

Water boils at about 91.7 °C on Mt. Whitney.

7 0

2 years ago

Write the equation of the chemical reaction of burning glucose (C6H12O6)

kotegsom [21]

Answer:

C6H12O6 + 6O2 → 6CO2 + 6H2O.

Explanation:

8 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago

Consider a balloon with volume V. It contains n moles of gas and has an internal pressure of P. The temperature of the gas is T.

Maksim231197 [3]

Answer:

V₂ = 0.6 V.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n is constant, and have different values of P, V and T:

(P₁V₁T₂) = (P₂V₂T₁).

V₁ = V, P₁ = P, T₁ = T.

V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.

∴ V₂ = (P₁V₁T₂)/(P₂T₁) = (P)(V)(0.75 T)/(1.25 P)(T) = 0.6 V.

3 0

3 years ago

Read 2 more answers

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

Ulleksa [173]

Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:

$0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY$

I hope it helps!

6 0

3 years ago