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irina [24]
2 years ago
14

Write the number 0.0000087 in scientific notation.

Chemistry
2 answers:
miskamm [114]2 years ago
6 0

Answer:

8.7 × 10-6

Explanation:

sorry if incorrect

natali 33 [55]2 years ago
3 0

Answer:

8.7 \times  {10}^{ - 6}

Explanation:

0.0000087

8.7 \times  {10}^{ - 6}

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What is the normality of a solution containing 14.8 g of Ca(OH)₂ in 250.0 mL?
aliina [53]

Answer:

Explanation:Are You From Milo?

6 0
3 years ago
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of ccl4. the b
Finger [1]
Answer : The molar mass of the solute will be 87.90 g/mol.

Explanation : We know the formula for elevation in boiling point, which is

Δt = iK_{b}m

given that, Δt = 0.357, K_{b} = 5.02 and mass of CCl _{4} = 40,

on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X10^{-3}. 
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3 0
3 years ago
What is the equilibrium constant of pure water at 25°C?
ANEK [815]
The equilibrium reaction, causes the water dissociation constant, Kw, is 1.01 × 10-14<span> at 25 °C. That is because every H</span>+<span> (H</span>3O+) ion these forms accompanied by the formation of an OH-<span> ion, are the concentrations of these ions and in pure water the same thing can be calculated from </span>Kw<span>. 

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6 0
3 years ago
Read 2 more answers
The volume of a weather balloon is 200.0 L and its internal pressure is 1.17 atm when it is launched at 20 °C. The balloon rises
brilliants [131]

Answer:

2022 L

Explanation:

Ideal gas laws will work for gas in the balloon

The general gas law is for a gas at two arbitrary states 1 and 2 is given by

(P₁ V₁)/T₁ = (P₂ V₂)/T₂

P₁ = 1.17 atm

V₁ = 200.0 L

T₁ = 20°C = 293.15 K

P₂ = 63 mmHg = 0.0829 atm

V₂ = ?

T₂ = 210 K

(1.17 × 200)/293.15 = (0.0829 × V₂)/210

V₂ = (210 × 1.17 × 200)/(293.15 × 0.0829)

V₂ = 2022 L

8 0
3 years ago
A student needs to prepare 50.0 mL of a 1.20 M aqueous H2O2 solution. Calculate the volume of 4.9 M H2O2 stock solution that sho
Nonamiya [84]

Answer : The volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

Solution : Given,

Molarity of aqueous H_2O_2 solution = 1.20 M = 1.20 mole/L

Volume of aqueous H_2O_2 solution = 50.0 ml = 0.05 L

(1 L = 1000 ml)

Molarity of H_2O_2 stock solution = 4.9 M = 4.9 mole/L

Formula used :

M_1V_1=M_2V_2

where,

M_1 = Molarity of aqueous H_2O_2 solution

M_2 = Molarity of H_2O_2 stock solution

V_1 = Volume of aqueous H_2O_2 solution

V_2 = Volume of H_2O_2 stock solution

Now put all the given values in this formula, we get the volume of H_2O_2 stock solution.

(1.20mole/L)\times (0.05L)=(4.9mole/L)\times V_2

By rearranging the term, we get

V_2=0.01224L=12.24ml

Therefore, the volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

3 0
3 years ago
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