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Lady_Fox [76]
2 years ago
6

4 features of bacteria that enable them to survive.

SAT
1 answer:
den301095 [7]2 years ago
4 0

Answer:

Specific structural features that enable prokaryotes to thrive in diverse environments include their cell walls (which provide shape and protection), flagella (which function in directed movement), and ability to form capsules or endospores (both of which can protect against harsh conditions). Prokaryotes also possess biochemical adaptations for growth in varied conditions, such as those that enable them to tolerate extremely hot or salty environments.

Explanation:

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Which equation is an example of the commutative property of multiplication? (4 2i) = (2i 4) (4 2i)(3 – 5i) = (3 – 5i)(4 2i) (4 2
kozerog [31]

Option B is correct. (4 + 2i)(3 – 5i) = (3 – 5i)(4 + 2i) satisfies the commutative property of multiplication

Given two values A and B, according to the commutative property of multiplication;

  • AB = BA

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From the given option, we can see that (4 + 2i)(3 – 5i) = (3 – 5i)(4 + 2i) satisfies the commutative property of multiplication where we can say;

A = 4 + 2i and B = 3 - 5i

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Calculate the mass of hydrogen in 57. 010 grams of ammonium hydrogen phosphate.
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The mass of hydrogen in 57.010 g ammonium hydrogen phosphate, (NH₄)H₂PO₄ is 2.97 g  

<h3>Determination of mass of 1 mole of (NH₄)H₂PO₄ </h3>

1 mole of (NH₄)H₂PO₄ = 14 + (4×1) + (2×1) + 31 + (16×4) = 115 g

<h3>Determination of the mass of H in 1 mole of (NH₄)H₂PO₄ </h3>

Mass of H = 6H = 6 × 1 = 6 g

<h3>Determination of the mass of H in 57.010 g of (NH₄)H₂PO₄ </h3>

115 g of (NH₄)H₂PO₄ contains 6 g of H.

Therefore,

57.010 g of (NH₄)H₂PO₄ will contain = (57.010 × 6) / 115 = 2.97 g of H

Thus, 2.97 g of Hydrogen, H is present in 57.010 g of (NH₄)H₂PO₄

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brainly.com/question/13531044

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