A sample of gold has a mass of 67.2 g and a volume of 3.5 cm3. the density of gold is

Which of the following shows a coefficient?

rosijanka [135]

Answer:

my answer is b oti d I am not sure

6 0

2 years ago

Extend your thinking: The slow decay of radioactive materials can be used to find the age of rocks, fossils, and archaeological

lisov135 [29]

The age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

<h3>What is the half-life of a radioactive material?</h3>

The half-life of a radioactive material is the time taken for half the atoms present in the material to decay or disintegrate.

The half-life, $t_\frac{1}{2}$ , the age, t, and amount remaining, $A_{r}$ , of a radioactive material are related by the formula below:

$t = \frac{t_\frac{1}{2}*A_{r}}{-ln2}$

Half-life of carbon-14 = 6000 years

For the Egyptian papyrus with 63% of its original carbon-14 atoms: $t = \frac{6000*0.63}{-0.63} = 4000\:years$

For the Aboriginal charcoal with 22% of its original carbon-14 atoms:

$t = \frac{6000*0.22}{-0.63} = 13106.5\:years$

For the Mayan headdress with 79% of its original carbon-14 atoms:

$t = \frac{6000*0.79}{-0.63} = 2040\:years$

Neanderthal skull with 3% of its original carbon-14 atoms:

$t = \frac{6000*0.03}{-0.63} = 30353\:years$

Therefore, the age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

Learn more about half-life at: brainly.com/question/26689704

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6 0

2 years ago

Antifreeze is added to a car's radiator to protect it. Which of the following

zhenek [66]

Answer:

the correct answer is b antifreeze protects the radiator during summer in winter.

Explanation:

Antifreeze is (typically mixed with water) to help manage the engine during extreme temperatures. As the temperature fluctuates between hot and cold coolant is distributed throughout the engine block to sustain even operating temperature.

5 0

3 years ago

Calculate the [oh−] in a solution with a ph of 12.52.

BaLLatris [955]

PH + pOH = 14

12.52 + pOH = 14

pOH = 14 - 12.52

pOH = 1.48

[OH⁻] = 10^ -pOH

[OH⁻] = 10 ^- 1.48

[OH⁻] = 0.033 M

6 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

3 years ago