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BabaBlast [244]
3 years ago
5

Why are nonnative species a threat to biodiversity?

Chemistry
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

A. They often use up resources that other organisms need.

Explanation:

Invasive alien species are animals, plants, fungi and microorganisms entered and established in the environment from outside of their natural habitat. T

hey reproduce rapidly, <u><em>out-compete native species for food, water and space</em></u>, and are one of the main causes of global biodiversity loss.

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Calcium dihydrogen phosphate, Ca(H₂PO₄)₂, and sodium hydrogen carbonate, NaHCO₃, are ingredients of baking powder that react to
NikAS [45]

0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

<h3>What is baking powder?</h3>
  • Baking powder is a dry chemical leavener composed of carbonate or bicarbonate and a weak acid.
  • The addition of a buffer, such as cornstarch, prevents the base and acid from reacting prematurely.
  • Baking powder is used in baked goods to increase volume and lighten the texture.

To find how many moles of CO₂ are produced from 1.00 g of baking powder:

The balanced equation is:

  • Ca(H₂PO₄)₂(s) + 2NaHCO₃(s) → 2CO₂(g) + 2H₂O(g) + CaHPO₄(s) + Na₂HPO₄(s)

On 3.50 g of baking power:

  • mCa(H₂PO₄)₂ = 0.35 × 3.50 = 1.225 g
  • mNaHCO₃ = 0.31 × 3.50 = 1.085 g

The molar masses are: Ca = 40 g/mol; H = 1 g/mol; P = 31 g/mol; O = 16 g/mol; Na = 23 g/mol; C = 12 g/mol.

So,

  • Ca(H₂PO₄)₂: 40 + 4 × 1 + 31 + 8 × 16 = 203 g/mol
  • NaHCO₃: 23 + 1 + 12 + 3 × 16 = 84 g/mol

The number of moles is the mass divided by molar mass, so:

  • nCa(H₂PO₄)₂ = 1.225/203 = 0.006 mol
  • nNaHCO₃ = 1.085/84 = 0.0129 mol

First, let's find which reactant is limiting.

Testing for Ca(H₂PO₄)₂, the stoichiometry is:

  • 1 mol of Ca(H₂PO₄)₂ ---------- 2 mol of NaHCO₃
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol

So, NaHCO₃ is in excess.

The stoichiometry calculus must be done with the limiting reactant, then:

  • 1 mol of Ca(H₂PO₄)₂ ------------- 2 mol of CO₂
  • 0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

  • x = 0.012 mol of CO₂

Therefore, 0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

Know more about baking powder here:

brainly.com/question/20628766

#SPJ4

The correct question is given below:

Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise: Ca(H2PO4)2(s) + NaHCO3(s) → CO2(g) + H2O(g) + CaHPO4(s) + Na2HPO4(s)[unbalanced] If the baking powder contains 31.0% NaHCO3 and 35.0% Ca(H2PO4)2 by mass: (a) How many moles of CO2 are produced from 3.50 g of baking powder?

3 0
2 years ago
What devices will you use to measure the mass and the volume in this experiment? (density)
bekas [8.4K]

Answer:

Methods for determining or delivering precise volumes include volumetric pipets and pycnometers; less precise methods include burets, graduated cylinders, and graduated pipets. In this experiment, you will measure masses and volumes to determine density. Four different metal cylinders are investigated.

Explanation:

8 0
2 years ago
when Jane drives to work, she always places her purse on the passengers seat. by the time she gets to work, her purse has fallen
marishachu [46]
Hmm, friction maybe? I guess it depends on how fast she stopped?
7 0
3 years ago
Determine the number of moles in 497 grams of the
satela [25.4K]

Answer:

• The actual number of moles of each element in the smallest unit of the compound. •In water (H 2 O), ammonia (NH 3), methane (CH 4), and ionic compounds, the empirical and molecular

Explanation:

5 0
2 years ago
A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is
Vlada [557]

Answer:

c=0.127\ J/g^{\circ} C

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

Q=mc\Delta T

c is specific heat of the metal

c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C

So, the specific heat of metal is 0.127\ J/g^{\circ} C.

4 0
2 years ago
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