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3 years ago

How did the discoveries of scandium (sc) and germanium (ge) affect mendeleevâs work?

Chemistry

2 answers:

lukranit [14]3 years ago

3 0

they showed mandeleeves predictions were correct

OlgaM077 [116]3 years ago

3 0

Answer:

they showed mandeleeves predictions were correct

Explanation:

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Explain how you would calculate the q for warming 100.0 grams of liquid water from 0°C to 100 °C.

mojhsa [17]

Answer:

mass = 100 g

T1 = 0°C

T2 = 100 °C

C = 1 cal/g°C

Q = mC(T2 -T1)

Q = 100(1)(100 - 0)

Q = 100(100)

Q = 10000 cal

Explanation:

6 0

2 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

2 years ago

If an acid has a ka=1.6x10^-10, what is the acidity of the solution?

LUCKY_DIMON [66]

The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+. If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).

5 0

2 years ago

Rewrite each equation below with the delta h value included with either the reactants or he products , and identify the reaction

grin007 [14]

I will take a stab at it, but there are not equations, did you forget them?

5 0

3 years ago

What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.

pav-90 [236]

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

Explanation:

ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

In our case we have:

mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if 12.41 g of MgBr₂ are dissolved in 2.55 L of water

then X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

then in 4.867 g of MgBr₂ we have Y g of Br⁻

Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)

4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

7 0

3 years ago