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3 years ago

8

The chart seen here shows the decay of a sample of carbon-14 over a period of 33,000 years. What is the approximate half-life of

carbon 14?

2 answers:

madreJ [45]3 years ago

7 0

A. 5,500 years is the answer to this question

Nadusha1986 [10]3 years ago

5 0

5,500 ears if u using usatestprep

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What is the concentration of the barium hydroxide solution if 50.0 mL of a 0.425 M HNO3 solution is required to neutralize a 36.

fomenos

The molarity of Barium Hydroxide is 0.289 M.

<u>Explanation:</u>

We have to write the balanced equation as,

Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O

We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,

V1M1 = 2 V2M2

Here V1 and M1 are the volume and molarity of nitric acid

V2 and M2 are the volume and molarity of Barium hydroxide.

So the molarity of Ba(OH)₂, can be found as,

$$ M2 = \frac{V1 \times M1}{2 \times V2}$

$$M2 = \frac{50 \times 0.425 }{2 \times 36.8}$

= 0.289 M

5 0

3 years ago

A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure

Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.

$P_i=P_t_o_t_a_l*X_i$

Where:

$P_i$ =Partial pressure

$P_t_o_t_a_l$ =Total pressure

$X_i$ =mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

<u>moles of hydrogen gas</u>

The molar mass of hydrogen gas ( $H_2$ ) is 2 g/mol, so:

$6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2$

<u>moles of oxygen gas</u>

The molar mass of oxygen gas ( $O_2$ ) is 32 g/mol, so:

$64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2$

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

$P_H_2=\frac{3}{5}*P_t_o_t_a_l$

$P_O_2=\frac{2}{5}*P_t_o_t_a_l$

So, the answer would be <u>3/5P</u>.

I hope it helps!

5 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

Volume of HCl used 25.0mL 4 l

borishaifa [10]

Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB = 25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

8 0

3 years ago

identify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet,

siniylev [52]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

3 0

3 years ago