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2 years ago

PLEASEEE HELP me on CHEMISTRY!! Will give brainliest!

Chemistry

1 answer:

Vlad [161]2 years ago

4 0

Answer:

A Glucose C6H12O6 is a covalent compound while NaCl sodium chloride is a ionic compound.

B Stronger molecular forces are present in NaCl because ionic bond is stronger than covalent bond.

C melting point allow us to predict the strength of intermolecular forces

High melting point stronger molecular forces and vice versa.

Explanation:

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Identify the for of energy (A)thermal (B)kinetic (C)chemical (D)electrical

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Answer:

C Chemical Energy

Explanation:

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A force of 450 N accelerates a box across a frictionless surface at a rate of 15 m/s2. What is the mass of the box?

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A reaction between liquid reactants takes place at −10.0°C in a sealed, evacuated vessel with a measured volume of 45.0L. Measur

maks197457 [2]

Answer: 0.0624 atm

Explanation:-

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = ?

V= Volume of the gas = 45.0 L

T= Temperature of the gas = -10.0°C = 263 K $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

Moles of gas= $\frac{\text{ given mass}}{\text{ molar mass}}= \frac{19.0g}{146g/mole}=0.130moles$

$P=\frac{nRT}{V}=\frac{0.130\times 0.0821\times 263}{45.0}=0.0624atm$

The pressure of sulfur hexafluoride gas in the reaction vessel after the reaction is 0.0624 atm

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2 years ago

You have 1.6 x10 21 atoms of oxygen, what is the mass of that number of atoms

iren [92.7K]

Answer: 166.023×1023

Explanation:

Mass of 6.023×1023 atoms of oxygen = gram atomic mass of oxygen.

Mass of 6.023×1023 atoms of oxygen = 16g

∴ Mass of 1 atom of oxygen =166.023×1023

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2 years ago

what volume of N2 is required to convert 5.0L of hydrogen gas to ammonia? assume that all gases are at the same temperature and

____ [38]

Answer:

Approximately $1.7\; {\rm L}$ .

Explanation:

Nitrogen ${\rm N_{2}}\, (g)$ reacts with hydrogen ${\rm H_{2}}\, (g)$ at a $1:3$ ratio to produce ammonia ${\rm NH_3}\, (g)$ :

${\rm N_{2}}\, (g) + 3\; {\rm H_{2}}\, (g) \to 2\; {\rm NH_{3}}\, (g)$ .

The ratio between the coefficient of ${\rm N_{2}}\, (g)$ and the coefficient of ${\rm H_{2}}\, (g)$ is:

$\begin{aligned}\frac{n({\rm N_{2}})}{n({\rm H_{2}})} = \frac{1}{3}\end{aligned}$ .

Under the ideal gas assumptions, the same ratio would apply to the volume of ${\rm N_{2}}\, (g)$ and ${\rm H_{2}}\, (g)$ in this reaction:

$\begin{aligned}\frac{V({\rm N_{2}})}{V({\rm H_{2}})} = \frac{n({\rm N_{2}})}{n({\rm H_{2}})} = \frac{1}{3}\end{aligned}$ .

$\begin{aligned}V({\rm N_{2}})= \frac{1}{3}\, V({\rm H_{2}})\end{aligned}$ .

Given that $V({\rm H_{2}}) = 5.0\; {\rm L}$ :

$\begin{aligned}V({\rm N_{2}}) &= \frac{1}{3}\, V({\rm H_{2}}) \\ &= \frac{1}{3}\times 5.0\; {\rm L} \\ &\approx 1.7\; {\rm L}\end{aligned}$ .

(Rounded to $2$ significant figures.)

5 0

1 year ago