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Tpy6a [65]
3 years ago
9

How many copper atoms are in 70g of copper

Chemistry
1 answer:
Elanso [62]3 years ago
4 0

Answer:

n = 6.634\times 10^{23}\,atoms

Explanation:

The total amount of atoms is found by multiplying Avogradro's Number and the number of moles:

n = \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)\cdot \left(\frac{70\,g}{63.546\,\frac{g}{mol} } \right)

n = 6.634\times 10^{23}\,atoms

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The Lewis dot model of a metal atom has two dots. Which of the following is true about the atom?
a_sh-v [17]

Answer:

It is present in the second group of the periodic table

Explanation:

7 0
3 years ago
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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

7 0
3 years ago
8. A 220 mL sample of helium gas is in a cylinder with a movable piston at 105 kPa and 275K. The piston
Harrizon [31]

The sample has a new pressure of 274kPa. If at 105 kPa and 275K, a 220 mL sample of helium gas is contained in a cylinder with a moving piston. The sample is pushed till it has a 95.0 mL volume and 310K .

The macroscopic characteristics of ideal gases are related by the ideal gas law (PV = nRT). A gas is considered to be perfect if its particles (a) do not interact with one another and (b) occupy no space (have no volume). Where P= pressure  V= volume and T = temperature.

From ideal gas equation

P₁V₁/T₁ =P₂V₂/T₂

105×220÷275 = P₂ ×95÷310

P₂= (105×220×310)÷(275×95)

P2= 7161000/26125

P2 = 274.105 kPa

Hence, the new pressure of helium gas is 274kPa

To know more about Ideas gas equation

brainly.com/question/28837405

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