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nata0808 [166]
3 years ago
6

. If 69.15% exists as 63Cu with an atomic mass of 62.92, then what is the atomic mass of the second isotope?

Chemistry
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer: 64.8

Explanation:Please see attachment for explanation

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If the mass of the object below is 28g, what is the density of the object below. Units are in cm below. Please round your answer
Murljashka [212]

Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

Volume of object = 3cm × 2cm ×5cm

Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

d = 28 g/ 30 cm³

d = 0.93 g/cm³

3 0
3 years ago
Which substance is one that cannot be separated or broken down into simpler substances by chemical means? (Element, pure)
Y_Kistochka [10]

Answer:

gold wire (Au)

Explanation:

A substance that cannot be separated or broken down into simpler substances by chemical means is an element.

Between the given options only gold wire is an element, Au.

A way of knowing that it is an element is noticing that its formula is a single symbol which corresponds to an element found in the periodic table, unlike a combination of said symbols (like NaCl or H₂O).

5 0
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Hannah is investigating how all of Earths systems can affect each other at any given moment. What is one way the cryosphere can
snow_lady [41]
The cryosphere can affect the atmosphere by letting the Earth get to warm. The cryosphere protects the Earth from getting to warm.
4 0
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10 pointsss Where are tsunamis likely to originate?
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Tsunamis are likely to originate in the pacific ocean
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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
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