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blsea [12.9K]
3 years ago
8

When you convert to you remove the placeholder zeros

Chemistry
1 answer:
Stella [2.4K]3 years ago
8 0
Scientific notations
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Element R and Element Q have the same number of valence electrons. These elements have similar chemical behavior, but element R
I am Lyosha [343]

Answer:

I dont now because I am in 6 class

7 0
3 years ago
Calculate the molality of a solution formed by adding 6.30 g NH4CL to 15.7 g of water
babymother [125]

Answer:

Molality = 7.5 mol/kg

Explanation:

Given data:

Mass of NH₄Cl = 6.30 g

Mass of water = 15.7 g (15.7/1000 =0.016 kg)

Molality = ?

Solution:

Formula of molality:

Molality = Moles of solute / mass of solvent in gram

Now we will first calculate the number of moles of solute( NH₄Cl )

Number of moles = mass/ molar mass

Molar mass of  NH₄Cl = 53.491 g/mol

Number of moles = 6.30 g/  53.491 g/mol

Number of moles =  0.12 mol

Now we will calculate the molality.

Molality = Moles of solute / mass of solvent in gram

Molality =  0.12 mol / 0.016 kg

Molality = 7.5 m

or        (m=mol/kg)

Molality = 7.5 mol/kg

6 0
3 years ago
How is the periodic law demonstrated in halogens
inn [45]
Halogens is defined as the group of 7 periodic table. As, every periodic table contains 7 valence electrons and they only need 1 more to complete an outer shell, that is why they are extremely reactive. And according to the law that recurring patterns of the properties of elements arise when they are arranged in order of increasing atomic number. As the halogen all act very similarly with each other in chemical reaction, it is true.
3 0
3 years ago
Read 2 more answers
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

2*n_{acid}=n_{base}

And in terms of concentrations and volumes:

2*M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the volume of acid:

V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL

Best regards.

5 0
3 years ago
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Protons are positive
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