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Debora [2.8K]
3 years ago
12

Can you explain how to convert 50pis to atm ?

Chemistry
2 answers:
Marianna [84]3 years ago
6 0

Explanation:

1 atm = 101,325 Pascals = 760 mm Hg = 760 torr = 14.7 psi. The prefix "kilo" means "1,000", so one kilopascal = 1,000 Pa. Therefore, 101.325 kPa = 1 atm = 760 torr and 100 kPa = 1 bar = 750 torr

Luba_88 [7]3 years ago
3 0
To convert psi to atm, multiply the psi value by 0.06804596379 or divide by 14.6959488.
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Use the following equation to answer question 1-4. Make sure you balance first.
worty [1.4K]
<h3>Answer:</h3>

5.2 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O

[Given] 10.4 mol HCl

<u>Step 2: Identify Conversions</u>

[RxN] 6 mol HCl = 3 mol H₂O

<u>Step 3: Stoichiometry</u>

  1. Set up:                             \displaystyle 10.4 \ mol \ HCl(\frac{3 \ mol \ H_2O}{6 \ mol \ HCl})
  2. Multiply/Divide:               \displaystyle 5.2 \ mol \ H_2O
4 0
3 years ago
Uranium (VIII) Sulfide formula
tresset_1 [31]

Answer:

US₂

Explanation:

Uranium sulfide (US₂)

Uranium atomic symbol = U

Sulfur atomic symbol = S

Uranium valency = +4

Sulfur valency = -2

So;

Uranium sulfide (US₂)

8 0
4 years ago
Among important crop plants, nitrogen-fixing root nodules are most commonly an attribute of A) corn. B) legumes. C) wheat. D) me
7nadin3 [17]
<span>Among important crop plants, nitrogen-fixing root nodules are most commonly an attribute of
B) legumes</span>
3 0
4 years ago
How do the hydrosphere and atmosphere interact? give an example
vladimir1956 [14]

Answer:

Evaporation

Explanation:

Water in the ocean, rivers, lakes, etc. is part of the hydrosphere, and when that water evaporates it enters the atmosphere

3 0
3 years ago
Calculate the density of the aluminum cylinder with a diameter 0f 1.3 cm weighing 18 grams. Height of the cylinder is 5.2 Cm. Fi
Masja [62]

Answer:

Percent error = 3.7%

Explanation:

Given data:

Density of Al cylinder = ?

Weight of cylinder = 18 g

Diameter = 1.3 cm

Height = 5.2 cm

Actual density of Al = 2.7 g/cm³

Percent error = ?

Solution:

First of all we will calculate the volume of cylinder through given formula.

V = πr²h

r = diameter /2

V = 22/7 × (0.65 cm)²× 5.2 cm

V = 22/7 × 0.4225cm²× 5.2 cm

V = 6.89 cm³

Now we will calculate the density.

d = m/v

d = 18 g/ 6.89 cm³

d = 2.6 g/cm³

Percent error:

Percent error = measured value - actual value /actual value × 100

Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³  × 100

Percent error = 3.7%

Negative sign shows that measured or experimental value is less than actual value.

6 0
3 years ago
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