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d1i1m1o1n [39]
3 years ago
6

PLEASE HELP

Chemistry
1 answer:
mixer [17]3 years ago
3 0

Answer:

CuSO4 + 2NH4OH -----> Cu(OH)2 + (NH4)2SO4

Explanation:

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You are conducting a plant growth experiment and need to generate a solution with growth factor, auxin or indole-3-acetic acid.
Yuki888 [10]

Answer:

7,01 g of indole-3-acetic acid

Explanation:

The milimolar concentration (mM) is defined as the ratio of milimoles per Liter of solution. 400mM means 400mmoles / L that is the same of <em>0,4mol / L</em>

100mL are <em>0,1L</em>. Using these values:

0,1L × (0,4mol / L ) = 0,04moles of indole-3-acetic acid.

As the MW of the molecule is 175,2 g/mol:

0,04mol × (175,2g / mol) = <em>7,01 g of indole-3-acetic acid</em>

<em></em>

Thus, <em>you need 7,01 g of indole-3-acetic acid to generate your solution</em>.

I hope it helps!

5 0
4 years ago
1. A 18 M solution of an acid that ionizes only slightly in solution would be termed___
kondor19780726 [428]
<span>A 18 M solution of an acid that ionizes only slightly in solution would be termed
concentrated and weak. The concentration of the acid is high. The acid which dissociates partially in water is a weak acid. 

</span><span>Calculate the [H^+] for the aqueous solution in which [OH^-] is 1 x10^-9. Is this solution acidic, basic or neutral. To determine [H+] use:

1x10^-14 = [OH-][H+]
solve for [H+]
[H+] = 1x10^-14/1x10^-9
        = 1x10^-5</span>
4 0
3 years ago
A container of carbon dioxide has a volume of 260 cm at a temperature of 22.0°C. If the
Oxana [17]

Answer:

V₂ =279.4 cm³

Explanation:

Given data:

Initial volume = 260 cm³

Initial temperature = 22.0°C

Final temperature = 44.0°C

Final volume = ?

Solution;

22.0°C (22+ 273 = 295k)

44.0°C(44+273 = 317k)

Formula:

According to Charles's law

V₁/T₁ = V₂/T₂

Now we will put the values in formula:

V₂ = V₁×T₂ / T₁

V₂ = 260 cm³ × 317k / 295k

V₂ = 82420 cm³. k  / 295k

V₂ =279.4 cm³

3 0
4 years ago
880.7 decameters<br> as hectometers
Butoxors [25]

Answer:??

Explanation:

4 0
3 years ago
2
Law Incorporation [45]

Answer:

(a) W

(b) X

(c) Y

Explanation:

Let's consider the following table with melting and boiling points for 4 substances.

Substance     Melting Point (°C)       Boiling Point (°C)

     W                       -7                                  60

     X                       660                              2500

     Y                        180                               1330

     Z                         115                                445

Let's consider that:

  • Below the melting point, a substance is solid.
  • Between the melting and the boiling point, a substance is liquid.
  • Above the boiling point, a substance is gas.

(a) Which substance is a gas at 100°C?

At 100 °C, W is above the boiling point

(b) Which substance is a liquid for the largest range of temperature?

The largest difference between the melting point and the boiling point is that of X: 2500 - 660 = 1840.

(c) Which substance is liquid at 1000 °C and a gas at 2000°C?​

Y is between the melting and the boiling point at 1000 °C and above the boiling point at 2000 °C.

3 0
3 years ago
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