All categories

3 years ago

By friends leaving this app

Chemistry

2 answers:

Lunna [17]3 years ago

3 0

Answer:

Whyyyyyyyyyy

Don't please...

Makovka662 [10]3 years ago

3 0

Whyyyyyyyyyy?????????

You might be interested in

Explain how science can help you make more informed decisions about what to eat.

kvasek [131]

So you wont die from eating too much great food! XD

6 0

3 years ago

Read 2 more answers

Does gases have a kinetic attraction? Yes or no.

satela [25.4K]

Answer:

Explanation:

There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.

The average kinetic energy of gas particles is proportional to the absolute temperature of the gas, and all gases at the same temperature have the same average kinetic energy.

7 0

3 years ago

What type of reaction occurs between Mg and HCl and explain why?

lyudmila [28]

Bonding is ionic for sure
Mg has lost 2e,s to become Mg2+
and there are 2 Cl-
the strong attraction between the Mg2+ and 2Cl- form the ionic bonding

6 0

3 years ago

an object with a larger mass requires greater force to accelerate than an object with a smaller mass true or false?

mr Goodwill [35]

Answer:

true

Explanation:

because of its weight

5 0

3 years ago

Read 2 more answers

During a routine check of the fluoride content of Gotham City\'s water supply, the following results were obtained from replicat

Aleksandr-060686 [28]

Answer:

The mean is $x=0.503\frac{mg}{L}$

The 90% confidence interval is:

$i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}]$

Explanation:

1. First organize the data:

$x_{1}=0.487$

$x_{2}=0.487$

$x_{3}=0.511$

$x_{4}=0.511$

$x_{5}=0.519$

As there are 5 data, the sample size (n) is n=5

2. Calculate the mean x:

The mean is calculated adding up all the data and divide them between the sample size.

$x=\frac{0.511+0.487+0.511+0.487+0.519}{5}$

$x=0.503\frac{mg}{L}$

3. Find 90% confidence interval.

The formula to find the confidence interval is:

$i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})]$ (Eq.1)

where x is the mean, d is the standard deviation and n is the sample size.

And

$1-\alpha=0.90$

$\alpha=0.10$

$\frac{\alpha}{2}=0.05$

$z_{0.05}=1.645$

4. Find the standard deviation

$d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}$

$d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}$

$d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}$

$d=\sqrt{2.24*10^{-4}}$

$d=0.015$

5. Replace values in (Eq.1):

$i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})]$

For the addition:

$i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.514$

For the subtraction:

$i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.492$

The 90% confidence interval is:

$i_{0.90}=[0.492,0.514]$

4 0

3 years ago

By friends leaving this app​

By friends leaving this app