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2 years ago

11

suggest two observations that you can make when Calcium reacts with water and write a balanced equation for the reaction of calc

ium with water

1 answer:

Kamila [148]2 years ago

8 0

• Bubbles of a colourless, odourless gas are evolved

• The solution turns red litmus blue

Ca + 2H_2O→ Ca(OH)_2 + H_2

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3. A mass of 0.15 ounces is equal to how many grams?

garik1379 [7]

Answer:

Option C = 4.25 g

Explanation:

Ounce and grams are unit of mass. Ounce is larger unit while gram is smaller unit. The one ounce is consist of 28.35 g or we can say that one ounce is equal to 28.35 g. In order to convert the given ounce value into grams the value is multiply with 28.35 g.

Given data:

Mass = 0.15 ounce

Mass in gram = ?

Solution:

One ounce is equal to 28.35 g, so

0.15 × 28.35 = 4.25 g

5 0

3 years ago

What will happen to the chemical equilibrium if MgCl2 is added?

vaieri [72.5K]

I think the answer is c.

7 0

2 years ago

Read 2 more answers

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

2 years ago

¿Las cargas eléctricas forman parte de los átomos?

shusha [124]

Answer:

si. en los atomos los protones y los electrones llevan cargas electricas

7 0

2 years ago

A 0.0100 mol sample of Ca(OH)2 requires 50.00 mL of aqueous HCl for neutralization according to the reaction below. What is the

harkovskaia [24]

Answer:Cm=o,4 (Mol/l)

Explanation:

Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + H2O(l)

0,01 0,02

CM=n/V

6 0

2 years ago