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3 years ago

Negative charge is placed in an eletric

Chemistry

1 answer:

leonid [27]3 years ago

7 0

I would i have think it is either a) east or c) west b/c couldn’t go by it self

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A cube with sides 3cm.Its mass is 54g. will it sink or float

zloy xaker [14]

Answer:

It will sink.

Explanation:

Density of water=1gm/cc (In CGS)

Volume of cube=l^3 =3^3= 27cc

Mass of cube =54gm

Density of cube=M/V =54/27 =2gm/cc

Here, density of cube is greater than that of water. So, the cube sinks.

7 0

3 years ago

calculate the number of grams of CH3COONa * 3H2O (sodium acetate tri-hydrate) needed to make 250.0 mL of a CH3COOH (acetic acid)

anastassius [24]

Answer:

10.88 g

Explanation:

We have:

[CH₃COOH] = 0.10 M

pH = 5.25

Ka = 1.80x10⁻⁵

V = 250.0 mL = 0.250 L

$M_{CH_{3}COONa*3H_{2}O} = molar \thinspace mass = 136 g/mol$

The pH of the buffer solution is:

$pH = pKa + log(\frac{[CH_{3}COONa*3H_{2}O]}{[CH_{3}COOH]})$ (1)

By solving equation (1) for [CH₃COONa*3H₂O] we have:

$log [CH_{3}COONa*3H_{2}O] = pH - pKa + log [CH_{3}COOH]$

$log [CH_{3}COONa*3H_{2}O] = 5.25 - (-log(1.80 \cdot 10^{-5})) + log (0.10) = -0.495$

$[CH_{3}COONa*3H_{2}O] = 10^{-0.495} = 0.32 M$

Hence, the mass of the sodium acetate tri-hydrate is:

$m = moles*M = [CH_{3}COONa*3H_{2}O]*V*M = 0.32 mol/L*0.250 L*136 g/mol = 10.88 g$

Therefore, the number of grams of CH₃COONa*3H₂O needed to make an acetic acid/sodium acetate tri-hydrate buffer solution is 10.88 g.

I hope it helps you!

4 0

3 years ago

Calculate the equilibrium constant at 298 K for the reaction of formaldehyde (CH2O) with hydrogen gas using the following inform

Mariana [72]

Answer:

E. 8.08 x 10⁴.

Explanation:

Hello,

In this case, for the reaction:

$CH_2O(g) + 2H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)$

We can compute the Gibbs free energy of reaction via:

$\Delta G\°=\Delta H\°-T\Delta S\°$

Since both the entropy and enthalpy of reaction are given at 298 K (standard temperature), therefore:

$\Delta G\°=-94.9kJ-(298K)(-224.2\frac{J}{K}*\frac{1kJ}{1000kJ} )\\\\\Delta G\°=-28.1kJ$

Then, as the equilibrium constant is computed as:

$K=exp(-\frac{\Delta G\°}{RT} )$

We obtain:

$K=exp(-\frac{-28.1kJ/mol}{8.314x10^{-3}\frac{kJ}{mol* K}}*298K )\\\\K=8.08 x10^4$

For which the answer is E. 8.08 x 10⁴.

Best regards,

3 0

4 years ago

Define independent variable

GenaCL600 [577]

A variable that stands alone and isn't changed by the other variables you are trying to measure. For example, if you’re trying to see if drinking milk affects a persons height, then the independent variable is the milk, and the dependent variable is the persons height change

4 0

2 years ago

Suppose three neutrons are released when an atom in a sample of fissionable nuclei undergoes fission. Each of these neutrons has

Eddi Din [679]

Answer:

13 nuclei

Explanation:

From the question, the fission of one nucleus produces three neutrons which causes more nuclei to undergo fission.

This implies that, after the first fission, three neutrons cause three nuclei to undergo fission. The three nuclei that underwent fission produces nine neutrons which causes the fission of nine nuclei.

All together we the number of nuclei that underwent fission as;

1 + 3 + 9 = 13 nuclei.

3 0

3 years ago