Pls help me with this question

How many equivalents are present in 10 g of Ca2+? <br> 1 <br> 0.5 <br> 1.5 <br> 2

Lynna [10]

0.5

Explanation:

Given parameters:

Mass of Ca²⁺ = 10g

unknown:

Equivalent weight = ?

Solution:

Equivalent weight that is the amount of electrons which a substance gains or loses per mole.

Ca²⁺ has +3 charge

It lost 2e⁻;

therefore;

In 1 mole of Ca²⁺, we have 2 equivalent weight

1 mol Ca²⁺ = 2eq. wts.

1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt.

Therefore;

10.0 g Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.

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Molar mass brainly.com/question/2861244

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8 0

3 years ago

Consider a hypothetical reaction in which A and B are reactants, and C and D are products. If 26 grams of A completely reacts wi

natali 33 [55]

Answer:

The answer to your question is: 24 grams of D

Explanation:

To answer this question we need to remember the Lavoisier law of conservation of mass, which says that in a chemical reaction matter is neither created nor destroyed.

This means that the amount of matter stays the same.

Then, the reaction is

A + B ⇒ C + D

26 g 12 g 14 g x

mass

of reactants 38 g ? mass of products, but it must be

equal to the mass of products

Then 14g + x = 38

x = 38 - 14

x = 24 g of D

4 0

3 years ago

12 *

nikitadnepr [17]

Answer:

true

Explanation:

7 0

3 years ago

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0

3 years ago

If 0.50 g of O2(g) reacts with excess H2(g), what is the volume of H2O(g) obtained from the reaction at STP?

ozzi

Answer:

0.7 L H2O

Explanation:

4 0

3 years ago