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Zanzabum
3 years ago
13

How many atoms of hydrogen are represented by the formula below? 4H2SO4 A. 1 B. 4 C. 8 D. 28

Chemistry
2 answers:
kirza4 [7]3 years ago
8 0
In this formula 8H atoms are represented
kkurt [141]3 years ago
3 0

2 atoms:

one mole is 6.022 times 10 to the 23 atoms according to Avogadro's Number, there are 2 Hydrogens in water, H20, so therefore 2 times 6.022 times to 10 to the 23 is your answer. O which means that ever molecule of water has 2 atoms of hydrogen (H) and one atom of oxygen (O)

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Its eeeeeeeeeeeezzzzzzzzzzzzzzzz science
artcher [175]

Answer:

B

Explanation:

Meters squared would be a good unit of measurement for Sanjeev's purposes.

4 0
3 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
What is the boiling point of a solution containing 203 g of ethylene glycol (C2H6O2) and 1035 g of water? (Kb for water is 0.52
Andru [333]

Answer:

101,37°C

Explanation:

Boiling point elevation is one of the colligative properties of matter. The formula is:

ΔT = kb×m <em>(1)</em>

Where:

ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-

kb is ebulloscopic constant (0,52°C/m)

And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):

203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>

<em />

Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m

Replacing these values in (1):

X - 100°C = 0,52°C/m×2,64m

X - 100°C = 1,37°C

<em>X = 101,37°C</em>

<em></em>

I hope it helps!

7 0
3 years ago
The lone pair of electrons on the oxygen atom in an ester can form hydrogen bonds in some situations, but esters cannot form hyd
mixer [17]

Answer:

Esters have lower boiling point than alcohols.

Explanation:

Esters are the fruity smelling compounds which are formed from carboxylic acid and alcohol with the removal of water.

The general formula for the ester is RCOOR' which is prepared from RCOOH acid and R'OH alcohol.

Ester can not form strong hydrogen bond as there is no hydrogen attached to the electronegative atom in the ester and thus cannot form hydrogen bonds with each other.<u> Due to this factor, the interactions within the molecules of the ester is lower than that of alcohols which exist in strong hydrogen bonding. As a result, ester can be easily boiled when compared to the alcohols and thus they have lower value of boiling points.</u>

4 0
3 years ago
Hello people ~
Natalija [7]

Answer:

Opposes the motion

Explanation:

Let's take an example

  • Suppose you kicked a ball and the ball is doing down the ground
  • The friction makes the ball to stop after sometime .
  • So it opposed the ball's motion

option A is correct

8 0
2 years ago
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