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butalik [34]
3 years ago
13

Consider the nuclear equation below. Superscript 22 subscript 11 upper N a right arrow superscript 22 subscript 10 upper N e plu

s superscript 0 subscript question mark beta. Which is the missing value that will balance the equation?
-1
0
+1
+2

Chemistry
2 answers:
natali 33 [55]3 years ago
8 0

Answer:

+1

Explanation:

For the equation to be balanced, the total mass number and the total atomic number on both side of the equation but be equal.

This is illustrated:

For the mass number:

Left side: 22

Right side: 22 + 0 = 22

For the atomic number:

Left side: 11

Right side: 10 + x

11 = 10 + x

Collect like terms

x = 11 - 10

x = 1

See attachment for further explanation.

Norma-Jean [14]3 years ago
4 0

Answer:

so basically its C

Explanation:

correct on edge 2020

if this helped rate 5 stars and press the heart

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The thin outer layer of Earth, called the crust, contains only 0.50 percent of Earth's total mass and yet is the source of almos
Rzqust [24]

Answer:

The mass of silicon in kilograms in Earth's crust is7.2729\times 10^{21} kg.

Explanation:

Mass of Earth =5.9\times 10^{21} tons

(1 ton= 2000 lb)

(1 lb =453.6 g)

1 ton = 2000 × 453.6 g =907,200 g

Mass of Earth =5.9\times 10^{21}\times 907,200 g=5.3524\times 10^{27} g

Percentage of earth crust = 0.50%

Mass of earth crust = M

0.50\%=\frac{M}{5.3524\times 10^{27}}\times 100

M=2.6762\times 10^{25} g

Percentage of the silicon in Earth's crust = 27.2 %

Mass of silicon in in Earth's crust = m

Si\%=\frac{m}{M}\times 100

27.2\%=\frac{m}{2.6762\times 10^{25} g}\times 100

m = 7.2792\times 10^{24} g=7.2729\times 10^{21} kg

1000 g = 1 kg

The mass of silicon in kilograms in Earth's crust is7.2729\times 10^{21} kg.

7 0
3 years ago
Suppose you need to prepare 136.9 mL of a 0.315 M aqueous solution of NaCl.
Varvara68 [4.7K]

Answer:

2.52 g NaCl

Explanation:

(Step 1)

To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:

Molarity = moles / volume (L)

After you convert mL to L, you can plug the given values into the equation and simplify to find moles.

136.9 mL / 1,000 = 0.1369 L

Molarity = moles / volume

0.315 M = moles / 0.1369 L

0.0431 = moles

(Step 2)

Now, you can use the molar mass to convert moles to grams.

Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol

Molar Mass (NaCl): 58.443 g/mol

0.0431 moles NaCl           58.443 g
------------------------------  x  -------------------  =  2.52 g NaCl
                                            1 mole

4 0
2 years ago
1. Addition of which of the following will increase the solubility of CaCO3 in water? Consider the equilibrium process:
kolbaska11 [484]

Answer: HCl

Explanation:

calcium carbonate dissolves in HCl acid producing CO 2 gas. It will not dissolve in pure water. The Ksp for calcium carbonate in water is 3.4 x 10-9 moldm-3 which is very low. What takes place here is actually a chemical reaction:

CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2CO 3(aq)

This reaction accounts for the solubility of the Calcium carbonate in HCl and not in pure water.

6 0
3 years ago
How many electrons occupy a filled 7s sublevel
Natali5045456 [20]

Answer:f has 14 electrons in 7 sublevel orbitals,d has 10 electrons in 5 sublevel orbitals,p has 6 electrons in 3 sublevel orbitals,s has 2 electrons in 1 sublevel orbital.

Explanation:

3 0
3 years ago
Read 2 more answers
What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use
ch4aika [34]

Explanation:

The given data is as follows.

          M_{1} = 10 mM = 10 \times 10^{-3} M

          V_{1} = 750 ml,           V_{2} = 5 ml

          M_{2} = ?

Therefore, calculate the molarity of given NaCl stock as follows.

                  M_{1} \times V_{1} = M_{2} \times V_{2}

                  10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml

                   M_{2} = 1.5 M

Thus, we can conclude that molarity of given NaCl stock is 1.5 M.        

7 0
3 years ago
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