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3 years ago

0.57 centimeter to meter

Chemistry

1 answer:

uysha [10]3 years ago

6 0

If 100cm is 1 metre
Then 0.57 cm =

If less more divides,
Hence 100÷0.57 × 1

= 0.0057 Metres.

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Find the volume of a gas at standard pressure if its volume at 1.9 atm is 80 ml?

kogti [31]

Answer:

1.5 × 10² mL

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 1.9 atm

Initial volume of the gas (V₁): 80 mL

Final pressure of the gas (P₂): 1.0 atm (standard pressure)

Final volume of the gas (V₂): ?

Step 2: Calculate the final volume of the gas

For an ideal gas, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1.9 atm × 80 mL/1.0 atm

V₂ = 1.5 × 10² mL

Since the pressure decreased, the volume of the gas increased.

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2 years ago

What is the pOH value of 0.0000877 M HCL

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Answer:

pOH=9.9

Explanation:

pH=-log[H+]= -log[0.0000877]

=4.06

pOH+ pH=14

pOH=14-4.06= 9.91

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3 years ago

You are measuring how many liters of oxygen are in a container. which of the following are you measuring?

svetoff [14.1K]

Answer:

Explanation:

liters is a measure of volume, it is an SI accepted metric system unit

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3 years ago

Mention one real life significance of the covalent bond.

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Answer:

it helps in respiration

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2 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

5 0

3 years ago