All categories

4 years ago

Please help!!!!!!!!!!!!!!

Chemistry

1 answer:

Elden [556K]4 years ago

4 0

Answer:

my guy we need to read lesson cuz int know what any of that mean

Explanation:

You might be interested in

What is the concentration of a dextrose solution prepared by diluting 16 mL of a 1.0 M dextrose solution to 25 mL using a 25 mL

seraphim [82]

Answer:

0.64 M

Explanation:

Step 1: Given data

Initial concentration (C₁): 1.0 M

Initial volume (V₁): 16 mL

Final concentration (C₂): ?

Final volume (V₂): 25 mL

Step 2: Calculate the final concentration of the dextrose solution

We want to prepare a diluted solution from a concentrated one. We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 1.0 M × 16 mL / 25 mL

C₂ = 0.64 M

4 0

3 years ago

It's easiest to weigh------using a balance scale?

allsm [11]

Yes and no it depends if your good at it you could also use a digital scale, but I perferr to use a balance scale

3 0

3 years ago

Please help me I need these answers

Cerrena [4.2K]

1 is true

2 is d 7

3 is

1 e

2 b

3 d

4 a

5 c

8 0

3 years ago

2.67 grams of butane (C4H10) is combusted in a bomb calorimeter. The temperature increases from 25.68 C to 36.2C. What is the ch

slavikrds [6]

The internal energy : 1310.43 kJ/mol

<h3>Further explanation</h3>

Internal energy (ΔE) can be formulated for Calorimeter :

$\tt \Delta E=C\times.\Delta t$

C= the heat capacity of the calorimeter

Δt=36.2-25.68=10.52°C

$\tt \Delta E=5.73 kJ/^oC\times 10.52^oC=60.2796~kJ$

mol butane(MW=58,12 g/mol)

$\tt \dfrac{2.67}{58,12 }=0.046$

the internal energy (ΔE) in KJ/mol

$\tt \dfrac{60.2796}{0.046}=1310.43~kJ/mol$

7 0

3 years ago

The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo

noname [10]

Answer:

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume.

Since we're finding the initial volume

$V_1 = \frac{P_2V_2}{P_1} \\$

We have

$V_1 = \frac{7.98 \times 14.2}{3.14} = \frac{113.316}{3.14} \\ = 36.0878...$

We have the final answer as

Hope this helps you

8 0

3 years ago