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I am Lyosha [343]
3 years ago
11

Someone please help me with my quiz about Moles for chemistry!

Chemistry
2 answers:
astra-53 [7]3 years ago
5 0

Answer:

A

C

A

Explanation:

Iam not sure if it's correct

lana66690 [7]3 years ago
3 0

ANSWER;

1: D

2: C

3: A

THANKS ME LATER

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What is the percent composition for the compound NaBr? (8C)
ehidna [41]

Answer:

the answer is D

Explanation:

percentage composition=  mole of the substance divided by the total molar mass of the compound multiplied by 100.

3 0
3 years ago
Explain why 1 mole of diethyl ether requires less heat to
gizmo_the_mogwai [7]

The quantity of heat required to vapourize 1 mole of a substance depends on the kind of intermolecular forces between the molecules of the substance. Diethyl ether molecules are held together by weak dispersion forces compared to the stronger hydrogen bonding in ethanol. Therefore,  1 mole of diethyl ether requires less heat to  vapourize than is required to vapourize 1 mole of  ethanol.

Intermolecular forces hold the molecules a substance together in a given state of matter. The properties of a substance such as boiling point, melting point etc are dependent on the nature of intermolecular forces holding the molecules of the substance.

Diethyl ether molecules are held together by weak dispersion forces while molecules of ethanol are held together by hydrogen bonds.

Since hydrogen bonds are much stronger than dispersion forces, a greater quantity of heat is required to break the intermolecular hydrogen bonds in ethanol in order to vapourize them than is required to vapourize diethyl ether.

Therefore, owing to stronger intermolecular forces between molecules of ethanol, less heat is required to vapourize than is required to vapourize 1 mole of  ethanol.

Learn more: brainly.com/question/9328418

6 0
3 years ago
What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
3 years ago
A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

7 0
3 years ago
Find the molarity of 131 g Ba(NO3)2 dissolved in 750.0 water
storchak [24]

Answer:

6.68 x 10^-4

Explanation:

131g ÷ 261.337g/mol = 0.5012685 moles

0.5012685 moles ÷ 750.0 liters =

0.5012685÷ 750.0=0.000668358

6.68 x 10^-4

8 0
2 years ago
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