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docker41 [41]
3 years ago
11

How is 8.2 x 10^4 - 6.3 x 10^3 written in scientific notation

Chemistry
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

= 7.57 × 104

(scientific notation)

= 7.57e4

(scientific e notation)

= 75.7 × 103

(engineering notation)

(thousand; prefix kilo- (k))

Explanation:

Just in case this is all of them

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Which best illustrates how particle arrangement changes when solids change into liquids
alexandr402 [8]
I would say if you would add colors it would be the best

8 0
3 years ago
Read 2 more answers
N the reaction Mg + 2HCl -> MgCl2 + H2 which element’s oxidation number does not change?
BartSMP [9]

Answer:- Oxidation number of Cl does not change as it is -1 on both sides.

Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).

Oxidation number  of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.

On product side, Mg oxidation number is +2 as the oxidation number of alkaline earth metals in their compounds is +2.

Two Cl are present in magnesium chloride, so if Mg is +2 then Cl is -1.

Oxidation number of H on product side is 0 as it is present in its elemental for, H_2 ,

So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.


4 0
3 years ago
How many moles of hydrogen gas will be produced when 12 g of Mg will react completely with excess of an acid according to the fo
lina2011 [118]

Answer:

0.49 mol

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCI ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 12 g of Mg

The molar mass of Mg is 24.31 g/mol.

12g \times \frac{1mol}{24.31g} = 0.49mol

Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg

The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.

7 0
3 years ago
Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o
pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0
3 years ago
An ideal gas occupies a volume V at an absolute temperature T. If the volume is halved and the pressure kept constant, what will
Kruka [31]

Answer:

It will be halve of T

Explanation:

V1 = V

T1 = T

V2 = ½V

T2 = x

V1/T1 = V2/T2

V/T = ½V/x

Vx = ½VT

2Vx = VT

2x = T

x = ½T

6 0
3 years ago
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