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3 years ago

What could you do in an experiment to get more accurate results?

Chemistry

2 answers:

marissa [1.9K]3 years ago

4 0

I believe it’s b, correct me if i’m wrong

alexandr1967 [171]3 years ago

4 0

It will be D because it is always good practice to take more than five recordings/results/observations so that if you realize that the results vary a bit, you can find the average.

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What is the element above the element in Group 3, Period 4?

Dennis_Churaev [7]

Scandium? Is that what you mean?

5 0

3 years ago

PH of 0,035M HCl? How to count if your calculator hasn't got -lg?

IrinaK [193]

Given information:

Concentration of HCl = 0.035 M

To determine:

pH of the solution

Explanation:

Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions

HCl → H+ + Cl-

Hence the concentration of H+ = Cl- = 0.035M

Now, pH measures the strength of H+ ions in a given solution. It is expressed as:

pH = -log[H+]

pH (HCl) = -log(0.035) = 1.46

Ans: pH of 0.035M HCl is 1.46

6 0

3 years ago

An experiment shows that a 250 −mL gas sample has a mass of 0.436 g at a pressure of 742 mmHg and a temperature of 27 ∘C.

icang [17]

Answer:

41.9 g/ mol hope that helps you out

Explanation:

d=p.m/ r.t

8 0

3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

3 years ago

Calculate the volume of a sample of mercury with a density of 14.6 g/mL and a mass of 1.00 g. The answer is assumed to be in mL.

Darya [45]

Answer:

0.0685 mL

Explanation:

To find the volume of the sample, divide the mass by the density.

(1.00 g)/(14.6 g/mL) = 0.0685 mL

3 0

3 years ago