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bogdanovich [222]
2 years ago
7

What is GMOs? ( Thanks btw )​

Chemistry
1 answer:
Sunny_sXe [5.5K]2 years ago
5 0

Answer:

living organisms whose genetic material has been artificially manipulated in a laboratory through genetic engineering

Explanation:

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solution of 0.160 MNaOH is used to titrate 21.0 mL of a solution of H2SO4:H2SO4(aq)+2NaOH(aq)→2H2O(l)+Na2SO4(aq)If 38.8 mL of th
stealth61 [152]

Answer:

0.0432 M H2SO4

Explanation:

First, we want to find the moles of MNaOH used. We know that Molarity x Liters = moles. 0.160M x 0.0210L = 0.00336 moles MNaOH

to find the moles of H2SO4, we can use a mol ratio.

0.00336mol MNaOH x (1Mol H2SO4 /2mol MNaOH)

= 0. 00168 mol H2SO4

I found the mol ratio by looking at the coefficients in front of the molecules I knew(MNaOH) and the molecule I needed to find(H2SO4)

then, to find Molarity, we do mol/Liters

0.00168 mol/ 0.0388L =. 0.0432 M H2SO4

You can convert mL to L by dividing by 1000

the significant figures of this problem is 3, so my final answer will also have 3 sig figs.

3 0
3 years ago
Explain how materials are suited for different uses based on their physical and chemical properties?
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What is solution in chemistry
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a solution in chemistry is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.
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3 years ago
A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu
OLEGan [10]

Answer: The final temperature of nickel and water is  25.2^{o}C.

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, (t_{i}) = 23^{o}C,      

  Final temp, (t_{f}) = ?,

  Specific heat of water = 4.184 J/g^{o}C,      

Now, we will calculate the heat energy as follows.

           q = mS \Delta t

              = 55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, t_{i} = 100^{o}C,

   Final temperature, t_{f} = ?

 Specific heat of nickel = 0.444 J/g^{o}C

Hence, we will calculate the heat energy as follows.

          q = mS \Delta t

             = 15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)      

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

              q_{water}(gain) = -q_{alloy}(lost)

55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C) = -(15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C))

       t_{f} = \frac{25.9^{o}C}{1.029}

                 = 25.2^{o}C

Thus, we can conclude that the final temperature of nickel and water is  25.2^{o}C.

6 0
3 years ago
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