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Fofino [41]
3 years ago
8

When would you want to be at the beach during a spring tide?

Chemistry
1 answer:
meriva3 years ago
8 0
I would want to be at the beach during the spring.
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Two forces are acting on a wheelbarrow.One force is pushing to the right and an equal force is pushing to the left.What can you
andreev551 [17]
B. It is not accelerating since equal and opposite forces cancel each other. Hope it helps
7 0
4 years ago
Read 2 more answers
A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte
QveST [7]

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of  solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm^{3} or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then\frac{19*1000}{550} will saturate 1L

= 34.545g  will saturate 1Litre

The solubility thus is 34.55g/L

 

5 0
3 years ago
Which is best metal to use in an alloy to increase its electrical conductivity
creativ13 [48]
<span>I believe that its
Cu</span>
7 0
3 years ago
A reaction mixture initially contains 0.140 MCO and 0.140 MH2O. What will be the equilibrium concentration of CO?
Allushta [10]

Answer:

0.013 M

Explanation:

From the question, we can make the following deductions; we are given mixture that contains two compounds, that is A and B, 0.140 M CO and 0.140 M H2O respectively. Then, we are asked to find the equilibrium concentration of Carbonmonoxide,CO.

So, the equation for the reaction is given below;

CO + H2O <-----------------> CO2 + H2.

Initially: we have 0.14M of CO, 0.14M of H2O and zero (0) concentration of CO2 and H2.

At time,t = CO =0.14 - x , H2O = 0.14 - x, CO2 and H2 = x.

The above reaction consist of the forward reaction and the backward reaction.

Therefore, the equilibrium Concentration of CO;

(Since we are giving that Kc = 102). Then, Kc=  [CO2][H2] ÷ [CO][H2O]. Where Kc is the equilibrium constant.

Therefore, 102 = [x^2] / [0.14 - x]^2.

==> 10.1= x/0.14 - x.

====> 0.141 - 10.1 x = x.

x + 10.1 x = 0.141.

===> 11.1 x = 0.141.

===> x = 0.141 ÷ 11.1.

===> x = 0.127 M .

Then, at time,t CO = 0.14 - x.

= 0.14 - 0.127 = 0.013 M

8 0
4 years ago
Read 2 more answers
A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate
Zarrin [17]

Answer: Empirical formula is C_2H_5O

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = \frac{330g}{12g/mol}=27.5moles

Moles of Hydrogen = \frac{69.5g}{1g/mol}=69.5moles

Moles of Oxygen = \frac{220.2g}{16g/mol}=13.76moles

<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2

\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5

\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1

<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = C_2H_5O

7 0
3 years ago
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