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emmainna [20.7K]
2 years ago
5

Can someone help its not hard. What are the cons of using a cup made of plastic?

Chemistry
1 answer:
slavikrds [6]2 years ago
7 0

Answer:

it could burn if a hot liquid is poured into it and it could break easily

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According to Bohr atomic model
nata0808 [166]

Answer:

A small positively charged nucleus surrounded by revolving negatively charged electrons in fixed orbits

7 0
3 years ago
Calculate the volume occupied by 272g
cupoosta [38]

The answer for the following problem is mentioned below.

<u><em>Therefore volume occupied by methane gas is  184.78 × 10^-3 liters </em></u>

Explanation:

Given:

mass of methane(CH_{4}) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of  methane(CH_{4}) = 16.0​  grams

We know;

According to the ideal gas equation,

<u><em>P × V = n × R × T</em></u>

here,

n = m÷M

n =272 ÷ 16

<u><em>n = 17 moles</em></u>

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

<u><em>Therefore volume occupied by methane gas is  184.78 × 10^-3 liters </em></u>

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6 0
2 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
Part C<br> How did Dr. Tierno find the answer to his question?
matrenka [14]
Sorry I cant I just need some points
7 0
3 years ago
Read 2 more answers
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
2 years ago
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