All categories

2 years ago

Electrical Safety Test Take Exam

Chemistry

1 answer:

Eduardwww [97]2 years ago

5 0

Answer:

true tell me if i am right ok

Explanation:

You might be interested in

Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a

Makovka662 [10]

Explanation:

Tollens' reagent is prepared by using two-step process : -

Step 1:

Silver oxide is formed by mixing aqueous silver nitrate with base like sodium hydroxide. The reaction is shown below as:

$AgNO_3 + NaOH &\rightarrow AgOH + NHO_3 \nonumber \\ 2AgOH &\rightarrow Ag_2O + H_2O$

Step 2

Ammonia solution is drop-wise added until all the silver oxide dissolves to form the reagent. The reaction is shown below as:

$Ag_2O + 4NH_3 + H_2O \rightarrow 2Ag(NH_3)_2^+ + 2OH^-$

3 0

3 years ago

Which is a major environmental concern for our oceans?

padilas [110]

Plastic pollution is the correct answer

3 0

2 years ago

Read 2 more answers

Hi ! Could someone help me out with this last part to my chemistry practice ? Thank you !

skelet666 [1.2K]

After 100years, sample is 250g

After 200 years, sample is 125g

After 300years, sample is 62.5 g

7 0

3 years ago

A 2.3 kg object has 15 ) of kinetic energy. Calculate its speed.

chubhunter [2.5K]

Answer:

<h3>The answer is option D</h3>

Explanation:

To find the speed given the kinetic energy and mass we use the formula

$v = \sqrt{ \frac{2KE}{m} } \\$

where

m is the mass

v is the speed

From the question

KE = 15 J

m = 2.3 kg

We have

$v = \sqrt{ \frac{2 \times 15}{2.3} } = \sqrt{ \frac{30}{2.3} } \\ = 3.61157559...$

We have the final answer as

Hope this helps you

7 0

3 years ago

A wooden tool is found at an archaeological site. Estimate the age of the tool using the following information: A 100 gram sampl

lord [1]

Answer:

2578.99 years

Explanation:

Given that:

100 g of the wood is emitting 1120 β-particles per minute

Also,

1 g of the wood is emitting 11.20 β-particles per minute

Given, Decay rate = 15.3 % per minute per gram

So,

Concentration left can be calculated as:-

C left = $[A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Also, Half life of carbon-14 = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time =?

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}$

$\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}$

$0.7320=e^{-0.000120968\times t}$

$ln\ 0.7320=-0.000120968\times t$

<u>t = 2578.99 years</u>

8 0

3 years ago