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Calculate how many grams of BeCl2 are required to produce 0.52 grams of MnCl2

Leokris [45]

Answer:

65.0cp

Explanation:

7 0

3 years ago

Rachel loves gardening. This summer, her 10 ft x 10 ft garden space is divided equally among tomatoes, lettuce, and strawberries

Ray Of Light [21]

Answer:

Rachel needs 0.52 kg of fertilizer.

Explanation:

Since the surface is a square of 10 ft of size its area is A = 10 ft x 10 ft = 1.0 x 10² ft². One third of that space is devoted to tomatoes ("space is divided equally among tomatoes, lettuce, and strawberries"), that is, 1/3 x 100 ft² = 33 ft².

We calculate the kilograms of fertilizer required we need to know 2 equivalences:

1 ac = 43,560 ft²

1 lb = 0.4535 kg

Then, we can use proportions to find out the amount of fertilizer used:

$33ft^{2} \frac{1ac}{43,560ft^{2} } .\frac{1,500lb}{1ac} .\frac{0.4535kg}{1lb} =0.52kg$

7 0

3 years ago

O2 + C3H2 + H2O + CO2 Reaction type?

Kobotan [32]

Answer:

Combustion is the reaction type if you meant to put a "=" in between C3H2 and H2O

3 0

4 years ago

What is the volume of 3.40 mol of gas at 33.3C and 22.2atm of pressure

Luda [366]

The volume of 3.40 mol of gas at 33.3 C and 22.2 atm of pressure is 3.85 liter of gas. This problem can be solved by using the PV=nRT or V= nRT/V equation which is the relation between the molar volume, the temperature, and the pressure of gas. In this formula, P is the pressure, R is the universal gas constant ( 0.0821 atm L/mol K), n is the molecule amount, V is the molecular volume, and T is the temperature. The temperature used in this formula must be in Kelvin, therefore we have to convert the Celcius temperature into Kelvin temperature (33.3 C = 306.45 K).
The calculation for the problem above: 3.4*0.0821*306.45/22.2 = 3.85 liter of gas.

5 0

3 years ago

I NEED HELP PLEASE, THANKS!

Crazy boy [7]

Answer:

Here's what I get.

Explanation:

1. Brønsted-Lowry theory

An acid is a substance that can donate a proton to another substance.

A base is a substance that can accept a proton from another substance.

2. pH of ammonia

The chemical equation is

$\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \text{OH}$^{-}$$

For simplicity, let's re-write this as

$\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$$

(a) Set up an ICE table.

B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹: 0.335 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.335 + x x x

$\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.335}{1.8 \times 10^{-5}} = 28 000 > 400\\\\x \ll 0.335$

(b) Solve for [OH⁻]

$\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}$

(c) Calculate the pOH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.46 \times 10^{-3}) = 2.61$

(d) Calculate the pH

pH = 14.00 - pOH = 14.00 - 2.61 = 11.39

3 0

4 years ago