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3 years ago

2. What type of bond holds together nitrogen atoms?

Chemistry

1 answer:

viva [34]3 years ago

3 0

Answer:option c

Explanation:

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The law of constant composition states: All atoms of a given element have a constant composition and are different than atoms of

Airida [17]

Answer:

Explanation:

The law proves C. For examples no matter how water you have it will always have a 1:2 ratio of oxygen to hydrogen. :)

7 0

2 years ago

Convert 3828 ml to L

crimeas [40]

Answer:

divide the volume value by 1000

So 3828/1000=3.828

3 0

3 years ago

24 POINTS!!!!!!!!!!

Oliga [24]

Your answer is going to be A, because it was shoved harder, it will go faster

8 0

3 years ago

Read 2 more answers

What is the molecular formula for C5H7 if its molar mass is 536 g/mol?

hodyreva [135]

Answer:

(C5H7)8

Explanation:

It's empirical formula is given as; C5H7

Molar mass of carbon(C) = 12 g/mol

Molar mass of hydrogen(H) = 1 g/mol

We are told that it's molar mass is 536 g/mol.

To find the molecular formula;

Molecular formula = n × empirical formula

Thus;

n = 536/((12 × 5) + (1 × 7))

n = 8

Thus;

Molecular formula = (C5H7)8

8 0

3 years ago

A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g

Lynna [10]

Answer:

$\%NaHCO_3=61.2\%$

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

$NaHCO_3+HA\rightarrow NaA+H_2O+CO_2$

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

$m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g$

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

$\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%$

Best regards.

5 0

3 years ago