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muminat
3 years ago
9

I really need help with my science homework

Chemistry
2 answers:
Kamila [148]3 years ago
8 0
SYMBOL, GROUP, PERIOD
He, Nobel Gas, 2
C, Nonmetal, 6
NA, Alkali metals, 11
S, non-metal, 16
Sn, Post-transition metal, 50
Ca, Alkali metals, 20
Li, alkali metals, 3
F, nonmetal, 9
K, alkali metal, 19
Mn, Transition metals, 25
P, nonmetal, 15
That's all I can do. Sorry. Its in order from top.
LuckyWell [14K]3 years ago
5 0
I can help, pm me and i will help you on there if that's alright?
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Umnica [9.8K]
LiOH is the right answer.
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A silicon atom has an atomic number of 14. What information does the atomic number tell you? (Choose all possible answers)
Bezzdna [24]

Answer:

Silicon atoms have 14 protons.

Silicon atoms will react with other atoms in order to gain stability.

Silicon atoms have 14 electrons.

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4 years ago
Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample
k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

<u>Step 1:</u> Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

<u>Step 2:</u> The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

<u>Step 3:</u> Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

<u>Step 4:</u> Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

<u>Step 5</u>: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0
3 years ago
A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the
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Answer : (C) Hafnium is the most likely identity of the given substance.

Solution :  Given,

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Volume of given substance (V) = 3.5 Cm^{3}

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Formula used :    

Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}

Now,put all the values in this formula, we get

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So, we conclude that the density of given substance (13.4 g/Cm^{3}) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/Cm^{3} respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

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3 0
3 years ago
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neonofarm [45]

Answer:

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3 years ago
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