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8_murik_8 [283]
3 years ago
10

Give three different between temporary and permanent change​

Chemistry
1 answer:
Morgarella [4.7K]3 years ago
4 0

Answer:

temporary is only for specific amount of time and permanent is suppost to mean forever

Explanation:

all i can say is not much can be permanent

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A compound containing Na, C, and O is found to have 1.06 mol Na, 0.528 mol C, and 1.59 mol O. What is the empirical formula of t
Effectus [21]

Given :

Moles of Na : 1.06

Moles of C : 0.528

Moles of O : 1.59

To Find :

The empirical formula of the compound.

Solution :

Dividing moles of each atom with the smallest one i.e 0.528 .

So,

Na : 1.06/0.528 = 2.007 ≈ 2

C : 0.528/0.528 = 1

O : 1.59/0.528 = 3.011 ≈ 3

Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.

So, empirical formula is Na_2CO_3 .

Hence, this is the required solution.

5 0
2 years ago
Which element is oxidized in the reaction below? fe(co)5 (l) + 2hi (g) fe(co)4i2 (s) + co (g) + h2 (g)?
DanielleElmas [232]
Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂<span> (s) is x. For whole compound, the charge is zero.

Sum of oxidation numbers in all elements = Charge of the compound.

Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
 x + 4*0 + 2*(-1) = 0
 x + 0 - 2            = 0
                   x     = +2
Hence the oxidation state of Fe in product </span>Fe(CO)₄I₂ (s) is +2.

Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
   y        = 0

Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.
5 0
3 years ago
The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.
Naddika [18.5K]

Answer:

12.6.

Explanation:

  • We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

  • It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

4 0
3 years ago
If the mass is 10g and the volume<br> is 2cm3 what is the density?
SSSSS [86.1K]

Answer:

The density is 5 g/cm3

Explanation:

The density (δ) is the ratio between the mass and the volume of a compound:

δ=m/v= 10 g/2 cm3= 5 g/cm3

3 0
3 years ago
What is the weight of a 90kg man standing on the moon, where gmoon = 1.64 m/s2?
Doss [256]

Answer:

The weigth of a 90kg man standing on the moon is <u><em>147.6 N (option C)</em></u>

Explanation:

Weight is called the action exerted by the force of gravity on the body.

The mass (amount of matter that a body contains) of an object will always be the same, regardless of where it is located. Instead, the weight of the object will vary according to the force of gravity acting on it.

The formula that allows you to calculate the weight of any body is:

W = m*g

where:

  • W = weight measured in N.
  • m = mass measured in kg.
  • g = acceleration of gravity measured in m/s². The acceleration of gravity g is the same for all objects that fall due to gravitational attraction, whatever their size or composition. For example, as an approximate value on Earth, g = 9.8 m/s².

In this case,  the mass m has a value of 90 kg and the gravity g has a value of 1.64 m/s², which is the value of the acceleration of gravity of the moon. Then:

W=90 kg* 1.64 m/s²

<u><em>W= 147.6 N</em></u>

Finally, <u><em>the weigth of a 90kg man standing on the moon is 147.6 N (option C)</em></u>

3 0
3 years ago
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