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3 years ago

Which atomic property is different in each isotope of an element?

1 answer:

6 0

Every isotope of an element has a different number of neutrons, which means that the atomic property which is different in each isotope of an element is mass number.
Mass number depends on the number of neutrons in an element.

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002 (part 1 of 2) 10.0 points

BARSIC [14]

Answer:

4) 1.5 mol

Explanation:

Well, the equation is already balanced and the mole to mole ratio of reactants and products are all 1. So if the limiting reactant is HCl and you have 1.5 mol, you do the mole to mole ratio with NaCl and since it is 1 to 1, there'd be 1.5 mol of NaCl.

6 0

2 years ago

A mileage test is conducted for a new car model. Thirty randomly selected cars are driven for a month and the mileage is measure

Leni [432]

Answer: $(27.81,\ 29.39)$

Explanation:

Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .

Significance level : $\alpha: 1-0.95=0.05$

Critical value: $z_{\alpha/2}=1.96$

Sample mean : $\overline{x}=28.6$

Standard deviation : $\sigma=2.2$

The formula to find the confidence interval is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}$

i.e. $28.6\pm 0.787259889321$

$\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)$

Hence, the 95% confidence interval for the mean mpg in the entire population of that car model = $(27.81,\ 29.39)$

4 0

3 years ago

This is what the map is for question 4

serious [3.7K]

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4 0

2 years ago

Copper(ii) chloride reacts with sodium nitrate to produce copper(ii) nitrate and sodium chloride.

zlopas [31]

Answer:

CuCl2 + 2NaNO3 - > Cu(NO3)2 + 2NaCl

5 0

2 years ago

When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio

8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

$V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL$

4 0

3 years ago