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ipn [44]
3 years ago
12

List 3 Metals in the periodic table.

Chemistry
2 answers:
musickatia [10]3 years ago
6 0
Chromium , silver, zinc...
OLga [1]3 years ago
5 0

Answer:

NUMBER SYMBOL ELEMENT

3 Li Lithium

4 Be Beryllium

11 Na Sodium

12 Mg Magnesium

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An astronomer observes an asteroid in the solar system. He notes that the asteroid is three times farther from the Sun than Eart
ozzi

Answer:

3 AU

Explanation:

The distance from the Earth to the Sun is known as 1 AU, or 1 Astronomical Unit. If an asteroid is three times this distance, it is 3 AU away.

7 0
3 years ago
A testable question is one that _____.
cestrela7 [59]

Answer:

One that “Can be answered by conducting an experiment”

Explanation:

8 0
3 years ago
Read 2 more answers
What is the oh−concentration and ph in each solution? (a) 0.225 m koh, (b) 0.0015 m sr(oh)2?
shepuryov [24]
Q1)
We have been given the OH⁻ concentration, therefore we first need to find the pOH value and then the pH value.
pOH = -log [OH⁻]
pOH = -log (0.225 M)
pOH =  0.65
pH + pOH = 14
pH = 14 - 0.65 = 13.35

Q2)
pOH = -log[OH⁻]
pOH = -log (0.0015 M)
pOH = 2.82
pH + pOH = 14
pH = 14 - 2.82 
pH = 11.18

7 0
3 years ago
How many moles are in 20.50 grams of NH3?
Tema [17]
Sorry, the correct answer is A. 3.4 x 10 - 24

Hope I helped ; )
6 0
3 years ago
Read 2 more answers
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
3 years ago
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