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Mamont248 [21]
3 years ago
11

This is usually used for school work but, I'm using it for something different. You see, I like this dude that I met over playst

ation (we've known each other for three years now) that I really like, and I don't know how to tell him! So.... Even though this isn't about school, I still need help. This is really stupid but...yeah
Chemistry
1 answer:
shusha [124]3 years ago
8 0

Answer:

The best way to admit you like him would be to act normal. Don't be someone your not and just admit to him that you like him. Yolu could do it while ya'll are talking but don't just suddenly spring it on him. If he is a true friend, your friendship will remain even if he doesn't like you back.

Explanation:

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
Simply parmanent tissue real life application please help it's for my project ​
WITCHER [35]
The simple tissues are parenchyma, sclerenchyma and collenchyma. Chlorenchyma is a parenchyma, having chloroplast. It is a simple permanent tissue, having chloroplast.
8 0
3 years ago
What does the answer of element 3Na2S2O3?
Fynjy0 [20]

Answer:

Explanation:

3× 22+2×32+ 6×16=

66+64+96=236

4 0
2 years ago
How could you distinguish octane from 1-octene by a simple<br> chemical test?
lara [203]

卂几丂山乇尺

As the name indicate that in 1-octene there is a double bond on the first position in the eight carbon atoms chain THUS to detect its presence, we use bromine water test;

Add bromine water to the test tube containing octane/octene if the brown colour of bromine water disappears then octene is present.

3 0
3 years ago
Read 2 more answers
If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

Nt = 4 g

So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.

8 0
3 years ago
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