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cluponka [151]
4 years ago
9

Muszę napisać wzór sumatryczny wysyłam załącznik

Chemistry
1 answer:
ioda4 years ago
7 0

Cześć, nie mówię po polsku, więc zrobiłem to w Tłumaczu Google, więc przepraszam, jeśli coś brzmi śmiesznie.

chlor (VII) i tlen - ten wzór to CI_{2} O_{7}, a tlenek chloru jest bezwodnikiem kwasu nadchlorowego.

węgiel i wodór - wzór na węgiel i wodór to CnH2n + 2), jest to związek organiczny, a niższa klasyfikacja jest taka, że jest to również węglowodór aromatyczny.

Mam nadzieję, że to pomoże, błogosławionego i cudownego dnia! :-)

-Cutiepatutie

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A student needs to make 0.855 L of a sodium bromide solution for an experiment. The concentration of the required solution is 0.
d1i1m1o1n [39]

Answer:

Explanation:

Molarity =  moles/liters of solution

moles    = molarity x Liters of solution

            =  0.350 m/L x 0.855 L

            =  0.299 mol

               mass = moles  x  molar mass    

            = 0.299 mol x 102.89 g/ms

            =  30.78 grams

4 0
3 years ago
Can y'all give me some examples of physical changes
gogolik [260]
Something that melts, something that changes shapes (for instance, play dough being squished), something that boils, something being mixed or dissolved (but only if it doesn't chemically react), etc. A physical change is a change of state of matter. 
4 0
3 years ago
When water vapor is cooled from 250°c to 150°c in a rigid tank the vapor just starts to?
viktelen [127]
The water vapor will begin to contract. The particles will begin to bond with each other to seek heat so they're contracting.
3 0
4 years ago
Which part of the atom is responsible for chemical bonding?​
jek_recluse [69]

Answer:

valence electrons

Explanation:

The electrons of an atom, in particular valence electrons, are involved in chemical bonding

7 0
3 years ago
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion,
Norma-Jean [14]

Answer:

m_{NaNH_2}=30.42gNaNH_2

m_{H_2}=0.783gH_2

Explanation:

Hello,

In this case, the reaction between sodium and ammonia is:

2Na+2NH_3\rightarrow 2NaNH_2+H_2

Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:

n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa

And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):

n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa

In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:

m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2

m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2

Best regards.

3 0
3 years ago
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